Examkin

Definite Integrals and Area Under Curves

27 soru

Tüm türler27Çoktan seçmeli13Sayısal14
Soru 1Soru →

What is the value of the definite integral ∫13(2x+1) dx\int_{1}^{3} (2x + 1) \, dx?

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Cevap: 10

Cevap

10
The antiderivative of 2x+12x + 1 is x2+xx^2 + x. Evaluating at the upper boundary x=3x = 3 gives 32+3=123^2 + 3 = 12, and at the lower boundary x=1x = 1 gives 12+1=21^2 + 1 = 2. Subtracting the lower bound result from the upper bound result yields 12−2=1012 - 2 = 10.

Adım Adım Çözüm

1
Find the indefinite integral (antiderivative) of 2x+12x + 1
∫(2x+1) dx=x2+x\int (2x + 1) \, dx = x^2 + x
Apply the power rule of integration: ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1}.
2
Evaluate the antiderivative at the upper limit x=3x = 3
32+3=9+3=123^2 + 3 = 9 + 3 = 12
Substitute x=3x = 3 into x2+xx^2 + x.
3
Evaluate the antiderivative at the lower limit x=1x = 1
12+1=1+1=21^2 + 1 = 1 + 1 = 2
Substitute x=1x = 1 into x2+xx^2 + x.
4
Subtract the lower limit value from the upper limit value
12−2=1012 - 2 = 10
According to the Fundamental Theorem of Calculus: ∫abf(x)dx=F(b)−F(a)\int_{a}^{b} f(x)dx = F(b) - F(a).

Anahtar Kavram

Definite Integrals of Linear Polynomials
Soru 2Soru →

The area of the region bounded by the parabola y=kx−x2y = kx - x^2 (where k>0k > 0) and the xx-axis between its xx-intercepts at x=0x = 0 and x=kx = k is equal to 3636 square units. What is the value of the positive constant kk?

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Cevap: 6

Cevap

The value of the positive constant kk is 6.
The area bounded by y=kx−x2y = kx - x^2 and the x-axis from x=0x = 0 to x=kx = k is obtained by integrating kx−x2kx - x^2, which yields k36\frac{k^3}{6}. Setting k36=36\frac{k^3}{6} = 36 gives k3=216k^3 = 216, whose cube root is k=6k = 6.

Adım Adım Çözüm

1
Set up the definite integral representing the area bounded by the curve and the x-axis between the intercepts x=0x = 0 and x=kx = k.
∫0k(kx−x2) dx=36\int_{0}^{k} (kx - x^2) \, dx = 36
The area under a curve y=f(x)y = f(x) above the x-axis from x=ax = a to x=bx = b is given by ∫abf(x) dx\int_{a}^{b} f(x) \, dx.
2
Find the antiderivative and evaluate it at the limits x=kx = k and x=0x = 0.
\left[ \frac{kx^2}{2} - \frac{x^3}{3} \right]_{0}^{k} = \left(\frac{k(k)^2}{2} - \frac{k^3}{3}\right) - 0 = \frac{k^3}{6}
Applying the integration power rule ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1} and simplifying k32−k33=k36\frac{k^3}{2} - \frac{k^3}{3} = \frac{k^3}{6}.
3
Set the evaluated expression equal to 36 and solve for kk.
\frac{k^3}{6} = 36 \implies k^3 = 216 \implies k = 6
Multiplying both sides by 6 yields k3=216k^3 = 216, and taking the cube root gives k=6k = 6.

Anahtar Kavram

Definite Integral and Area Under Curve
Soru 3Soru →

Find the positive constant pp such that the value of the definite integral ∫0p(3x2+2x) dx\int_{0}^{p} (3x^2 + 2x) \, dx is equal to 1212.

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Cevap: 2

Cevap

The positive constant pp is equal to 22.
Evaluating the definite integral ∫0p(3x2+2x) dx\int_{0}^{p} (3x^2 + 2x) \, dx gives p3+p2p^3 + p^2. Setting this expression equal to 1212 forms the cubic equation p3+p2−12=0p^3 + p^2 - 12 = 0. Solving for the positive real solution yields p=2p = 2, since 23+22=8+4=122^3 + 2^2 = 8 + 4 = 12.

Adım Adım Çözüm

1
Find the antiderivative of the integrand
\int (3x^2 + 2x) \, dx = x^3 + x^2 + C
Applying the power rule for integration ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} term by term gives 3x33+2x22=x3+x2\frac{3x^3}{3} + \frac{2x^2}{2} = x^3 + x^2.
2
Evaluate the antiderivative at the limits 00 and pp
\left[ x^3 + x^2 \right]_0^p = (p^3 + p^2) - (0^3 + 0^2) = p^3 + p^2
Using the Fundamental Theorem of Calculus: substitute the upper limit pp and subtract the value at the lower limit 00.
3
Set up the polynomial equation using the given integral value
p^3 + p^2 = 12 \implies p^3 + p^2 - 12 = 0
The problem states that the definite integral equals 12.
4
Solve for the positive real value of pp
p = 2
Testing p=2p=2: 23+22=8+4=122^3 + 2^2 = 8 + 4 = 12. Factoring gives (p−2)(p2+3p+6)=0(p-2)(p^2 + 3p + 6) = 0, where the quadratic factor has a negative discriminant 32−4(1)(6)=−15<03^2 - 4(1)(6) = -15 < 0, making p=2p=2 the only real solution.

Anahtar Kavram

Definite integration with an unknown limit of integration
Soru 4Soru →

What is the value of the definite integral ∫0π3cos⁡(2x) dx\int_{0}^{\frac{\pi}{3}} \cos(2x) \, dx?

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Cevap: 34\frac{\sqrt{3}}{4}

Cevap

34\frac{\sqrt{3}}{4}
The antiderivative of cos⁡(2x)\cos(2x) is 12sin⁡(2x)\frac{1}{2}\sin(2x). Substituting the limits gives 12sin⁡(2π3)−12sin⁡(0)=12(32)−0=34\frac{1}{2}\sin\left(\frac{2\pi}{3}\right) - \frac{1}{2}\sin(0) = \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) - 0 = \frac{\sqrt{3}}{4}, which is correct.

Adım Adım Çözüm

1
Find the antiderivative of cos⁡(2x)\cos(2x)
∫cos⁡(2x) dx=12sin⁡(2x)+C\int \cos(2x) \, dx = \frac{1}{2}\sin(2x) + C
Using standard trigonometric integration rules, ∫cos⁡(kx) dx=1ksin⁡(kx)+C\int \cos(kx) \, dx = \frac{1}{k}\sin(kx) + C.
2
Substitute the upper limit x=π3x = \frac{\pi}{3} and lower limit x=0x = 0
[12sin⁡(2x)]0π3=12sin⁡(2π3)−12sin⁡(0)\left[\frac{1}{2}\sin(2x)\right]_{0}^{\frac{\pi}{3}} = \frac{1}{2}\sin\left(\frac{2\pi}{3}\right) - \frac{1}{2}\sin(0)
Apply the Fundamental Theorem of Calculus: ∫abf(x)dx=F(b)−F(a)\int_{a}^{b} f(x)dx = F(b) - F(a).
3
Evaluate the trigonometric values and simplify
12(32)−0=34\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) - 0 = \frac{\sqrt{3}}{4}
sin⁡(2π3)=sin⁡(120∘)=32\sin\left(\frac{2\pi}{3}\right) = \sin(120^\circ) = \frac{\sqrt{3}}{2} and sin⁡(0)=0\sin(0) = 0.

Anahtar Kavram

Definite Integration of Trigonometric Functions
Soru 5Soru →

What is the value of the definite integral ∫12(4x3−3x2+2) dx\int_{1}^{2} (4x^3 - 3x^2 + 2) \, dx?

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Cevap: 10

Cevap

10
Integrating each term yields F(x)=x4−x3+2xF(x) = x^4 - x^3 + 2x. Substituting the upper boundary gives F(2)=12F(2) = 12 and the lower boundary gives F(1)=2F(1) = 2. Calculating F(2)−F(1)F(2) - F(1) gives 12−2=1012 - 2 = 10, which is the exact value of the definite integral.

Adım Adım Çözüm

1
Find the indefinite antiderivative of the integrand 4x3−3x2+24x^3 - 3x^2 + 2.
F(x)=∫(4x3−3x2+2) dx=x4−x3+2xF(x) = \int (4x^3 - 3x^2 + 2) \, dx = x^4 - x^3 + 2x
Apply the power rule of integration ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} to each term.
2
Evaluate the antiderivative at the upper limit x=2x = 2.
F(2)=(2)4−(2)3+2(2)=16−8+4=12F(2) = (2)^4 - (2)^3 + 2(2) = 16 - 8 + 4 = 12
Substitute x=2x = 2 into F(x)F(x).
3
Evaluate the antiderivative at the lower limit x=1x = 1.
F(1)=(1)4−(1)3+2(1)=1−1+2=2F(1) = (1)^4 - (1)^3 + 2(1) = 1 - 1 + 2 = 2
Substitute x=1x = 1 into F(x)F(x).
4
Subtract the lower limit value from the upper limit value.
F(2)−F(1)=12−2=10F(2) - F(1) = 12 - 2 = 10
Apply the Fundamental Theorem of Calculus: ∫abf(x)dx=F(b)−F(a)\int_{a}^{b} f(x) dx = F(b) - F(a).

Anahtar Kavram

Definite Integration of Polynomial Functions
Tahmini Süre:1m 30s
Soru 6Soru →

Find the value of the definite integral ∫12(x+1x2)dx\int_{1}^{2} \left(x + \frac{1}{x^2}\right) dx.

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Cevap: 2

Cevap

The value of the definite integral is 2.
Integrating x+x−2x + x^{-2} gives x22−1x\frac{x^2}{2} - \frac{1}{x}. Evaluating from x=1x=1 to x=2x=2 yields (2−12)−(12−1)=32−(−12)=2\left(2 - \frac{1}{2}\right) - \left(\frac{1}{2} - 1\right) = \frac{3}{2} - \left(-\frac{1}{2}\right) = 2.

Adım Adım Çözüm

1
Find the antiderivative of f(x)=x+x−2f(x) = x + x^{-2}
F(x)=x22−1x+CF(x) = \frac{x^2}{2} - \frac{1}{x} + C
Apply the power rule of integration ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} for n≠−1n \neq -1.
2
Evaluate F(x)F(x) at the upper boundary x=2x = 2
F(2)=222−12=2−12=32F(2) = \frac{2^2}{2} - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2}
Substitute x=2x = 2 into the antiderivative.
3
Evaluate F(x)F(x) at the lower boundary x=1x = 1
F(1)=122−11=12−1=−12F(1) = \frac{1^2}{2} - \frac{1}{1} = \frac{1}{2} - 1 = -\frac{1}{2}
Substitute x=1x = 1 into the antiderivative.
4
Calculate F(2)−F(1)F(2) - F(1)
32−(−12)=32+12=2\frac{3}{2} - \left(-\frac{1}{2}\right) = \frac{3}{2} + \frac{1}{2} = 2
Apply the Fundamental Theorem of Calculus: ∫abf(x)dx=F(b)−F(a)\int_{a}^{b} f(x) dx = F(b) - F(a).

Anahtar Kavram

Definite Integration using the Power Rule
Soru 7Soru →

Evaluate the definite integral ∫02(3x2+4) dx\int_{0}^{2} (3x^2 + 4) \, dx.

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Cevap: 16

Cevap

The value of the definite integral is 1616.
Integrating 3x2+43x^2 + 4 with respect to xx gives the antiderivative F(x)=x3+4xF(x) = x^3 + 4x. Evaluating this antiderivative at the upper limit x=2x = 2 yields 23+4(2)=162^3 + 4(2) = 16, and at the lower limit x=0x = 0 yields 03+4(0)=00^3 + 4(0) = 0. Subtracting the lower boundary value from the upper boundary value gives 16−0=1616 - 0 = 16.

Adım Adım Çözüm

1
Integrate the polynomial term by term
\int (3x^2 + 4) dx = x^3 + 4x
Apply the power rule of integration \int x^n dx = \frac{x^{n+1}}{n+1} and \int k dx = kx.
2
Apply the fundamental theorem of calculus with limits 0 and 2
[x^3 + 4x]_0^2 = (2^3 + 4(2)) - (0^3 + 4(0)) = 16 - 0 = 16
Evaluate F(b) - F(a) where F(x) is the antiderivative.

Anahtar Kavram

Definite Integral Evaluation using the Fundamental Theorem of Calculus
Soru 8Soru →

Determine the area under the curve y=3x2−3y = 3x^2 - 3 above the xx-axis between x=1x = 1 and x=3x = 3.

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Cevap: 20

Cevap

The area under the curve between x=1x = 1 and x=3x = 3 is 20 square units.
The area under the curve y=3x2−3y = 3x^2 - 3 from x=1x = 1 to x=3x = 3 is obtained by calculating the definite integral ∫13(3x2−3) dx=[x3−3x]13=(33−3(3))−(13−3(1))=18−(−2)=20\int_{1}^{3} (3x^2 - 3) \, dx = [x^3 - 3x]_{1}^{3} = (3^3 - 3(3)) - (1^3 - 3(1)) = 18 - (-2) = 20.

Adım Adım Çözüm

1
Set up the definite integral representing the area under the curve.
A=∫13(3x2−3) dxA = \int_{1}^{3} (3x^2 - 3) \, dx
The area bounded by a non-negative curve y=f(x)y = f(x), the xx-axis, and vertical lines x=ax = a and x=bx = b is given by ∫abf(x) dx\int_{a}^{b} f(x) \, dx.
2
Integrate the polynomial function term by term.
∫(3x2−3) dx=x3−3x+C\int (3x^2 - 3) \, dx = x^3 - 3x + C
Applying the power rule of integration: ∫3x2dx=x3\int 3x^2 dx = x^3 and ∫3dx=3x\int 3 dx = 3x.
3
Apply the Fundamental Theorem of Calculus by substituting the limits of integration.
[x3−3x]13=(33−3(3))−(13−3(1))=18−(−2)=20[x^3 - 3x]_{1}^{3} = (3^3 - 3(3)) - (1^3 - 3(1)) = 18 - (-2) = 20
Evaluating F(b)−F(a)F(b) - F(a) gives (27−9)−(1−3)=18−(−2)=20(27 - 9) - (1 - 3) = 18 - (-2) = 20.

Anahtar Kavram

Definite Integrals and Area Under Curves
Soru 9Soru →

Given that ∫0p(6x−4) dx=15\int_{0}^{p} (6x - 4) \, dx = 15 and p>0p > 0, what is the value of pp?

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Cevap: 33

Cevap

The value of pp is 33.
Integrating 6x−46x - 4 yields 3x2−4x3x^2 - 4x. Substituting the limits from 00 to pp gives 3p2−4p3p^2 - 4p. Equating this to 1515 produces the quadratic equation 3p2−4p−15=03p^2 - 4p - 15 = 0, which factors as (3p+5)(p−3)=0(3p + 5)(p - 3) = 0. Since p>0p > 0, the only valid solution is 33.

Adım Adım Çözüm

1
Find the indefinite integral of the integrand f(x)=6x−4f(x) = 6x - 4.
∫(6x−4) dx=3x2−4x\int (6x - 4) \, dx = 3x^2 - 4x
Apply the power rule of integration: ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}.
2
Evaluate the definite integral from lower limit 00 to upper limit pp.
[3x2−4x]0p=(3p2−4p)−(0)=3p2−4p[3x^2 - 4x]_{0}^{p} = (3p^2 - 4p) - (0) = 3p^2 - 4p
Substitute the upper and lower limits into the antiderivative.
3
Set the evaluated definite integral equal to the given value of 1515 and form a quadratic equation.
3p2−4p−15=03p^2 - 4p - 15 = 0
Equate the definite integral value to 15.
4
Factor the quadratic equation to solve for pp.
(3p+5)(p−3)=0  ⟹  p=−53(3p + 5)(p - 3) = 0 \implies p = -\frac{5}{3} or p=3p = 3
Find two numbers that multiply to −45-45 and add up to −4-4, which are −9-9 and 55.
5
Select the valid positive value of pp as specified in the question (p>0p > 0).
p=3p = 3
Disregard the negative root since p>0p > 0.

Anahtar Kavram

Determining an unknown boundary limit of a definite integral
Soru 10Soru →

Evaluate the definite integral ∫13(4x3−6x) dx\int_{1}^{3} (4x^3 - 6x) \, dx.

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Cevap: 56

Cevap

The value of the definite integral is 56.
To evaluate the definite integral ∫13(4x3−6x) dx\int_{1}^{3} (4x^3 - 6x) \, dx, integrate the function term-by-term to obtain F(x)=x4−3x2F(x) = x^4 - 3x^2. Evaluating F(x)F(x) at the upper bound x=3x = 3 yields 5454, and evaluating at the lower bound x=1x = 1 yields −2-2. Subtracting the lower bound value from the upper bound value gives 54−(−2)=5654 - (-2) = 56.

Adım Adım Çözüm

1
Find the indefinite integral of the function
\int (4x^3 - 6x) \, dx = x^4 - 3x^2
Apply the power rule of integration \int x^n \, dx = \frac{x^{n+1}}{n+1}
2
Substitute the upper limit x = 3 into the antiderivative F(x)
F(3) = 3^4 - 3(3)^2 = 81 - 27 = 54
Evaluate F(b) for b = 3
3
Substitute the lower limit x = 1 into the antiderivative F(x)
F(1) = 1^4 - 3(1)^2 = 1 - 3 = -2
Evaluate F(a) for a = 1
4
Subtract F(a) from F(b)
54 - (-2) = 56
Apply the Fundamental Theorem of Calculus: \int_{a}^{b} f(x) \, dx = F(b) - F(a)

Anahtar Kavram

Definite Integration using the Fundamental Theorem of Calculus
Soru 11Soru →

Find the area of the region bounded by the curve y=3x2y = 3x^2, the xx-axis, and the vertical lines x=1x = 1 and x=3x = 3.

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Cevap: 26

Cevap

The area of the bounded region is 26 square units.
The area under y=3x2y = 3x^2 from x=1x = 1 to x=3x = 3 is calculated using the definite integral ∫133x2 dx=[x3]13=33−13=27−1=26\int_{1}^{3} 3x^2 \, dx = [x^3]_{1}^{3} = 3^3 - 1^3 = 27 - 1 = 26 square units.

Adım Adım Çözüm

1
Set up the definite integral representing the bounded area.
A=∫133x2 dxA = \int_{1}^{3} 3x^2 \, dx
The area under a non-negative curve y=f(x)y = f(x) from x=ax = a to x=bx = b above the xx-axis is given by the definite integral ∫abf(x) dx\int_{a}^{b} f(x) \, dx.
2
Determine the antiderivative of 3x23x^2.
∫3x2 dx=3⋅x33=x3\int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3
Applying the power rule of integration ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1} gives x3x^3.
3
Evaluate the definite integral using the fundamental theorem of calculus.
[x3]13=33−13=27−1=26[x^3]_{1}^{3} = 3^3 - 1^3 = 27 - 1 = 26
Substitute the upper limit x=3x = 3 and subtract the value of the function evaluated at the lower limit x=1x = 1.

Anahtar Kavram

Area under a curve using definite integration
Soru 12Soru →

What is the value of the definite integral ∫04(x+1) dx\int_{0}^{4} (x + 1) \, dx?

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Cevap: 12

Cevap

The value of the definite integral is 12.
Integrating x+1x + 1 yields x22+x\frac{x^2}{2} + x. Substituting the upper limit x=4x = 4 gives 422+4=8+4=12\frac{4^2}{2} + 4 = 8 + 4 = 12. Substituting the lower limit x=0x = 0 gives 00. Calculating 12−012 - 0 gives the final answer of 12.

Adım Adım Çözüm

1
Find the antiderivative of the integrand f(x)=x+1f(x) = x + 1.
∫(x+1) dx=x22+x+C\int (x + 1) \, dx = \frac{x^2}{2} + x + C
Apply the power rule of integration ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1} to x1x^1 and 1=x01 = x^0.
2
Evaluate the antiderivative at the upper bound x=4x = 4 and lower bound x=0x = 0.
F(4)=422+4=162+4=12F(4) = \frac{4^2}{2} + 4 = \frac{16}{2} + 4 = 12, and F(0)=022+0=0F(0) = \frac{0^2}{2} + 0 = 0
Substitute the bounds into the antiderivative expression F(x)=x22+xF(x) = \frac{x^2}{2} + x.
3
Apply the Fundamental Theorem of Calculus F(b)−F(a)F(b) - F(a).
F(4)−F(0)=12−0=12F(4) - F(0) = 12 - 0 = 12
Subtract the lower bound result from the upper bound result.

Anahtar Kavram

Evaluation of definite integrals of simple linear polynomials

Alternatif Yöntem

The integral ∫04(x+1) dx\int_{0}^{4} (x + 1) \, dx represents the area of a trapezoid under y=x+1y = x + 1 from x=0x = 0 to x=4x = 4. The parallel vertical sides are h1=1h_1 = 1 and h2=5h_2 = 5, with width b=4b = 4. Using the area formula for a trapezoid: Area=1+52×4=3×4=12\text{Area} = \frac{1+5}{2} \times 4 = 3 \times 4 = 12.
Tahmini Süre:45s
Soru 13Soru →

What is the value of the definite integral ∫0π6cos⁡(3x) dx\int_{0}^{\frac{\pi}{6}} \cos(3x) \, dx?

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Cevap: 13\frac{1}{3}

Cevap

The value of the definite integral is 13\frac{1}{3}.
The antiderivative of cos⁡(3x)\cos(3x) is 13sin⁡(3x)\frac{1}{3}\sin(3x). Substituting the upper boundary x=π6x = \frac{\pi}{6} yields 13sin⁡(π2)=13\frac{1}{3}\sin\left(\frac{\pi}{2}\right) = \frac{1}{3}, and substituting the lower boundary x=0x = 0 yields 00. Subtracting the lower bound evaluation from the upper bound evaluation gives 13\frac{1}{3}.

Adım Adım Çözüm

1
Find the indefinite integral of cos⁡(3x)\cos(3x)
∫cos⁡(3x) dx=13sin⁡(3x)+C\int \cos(3x) \, dx = \frac{1}{3}\sin(3x) + C
Using the standard integration rule ∫cos⁡(kx) dx=1ksin⁡(kx)+C\int \cos(kx) \, dx = \frac{1}{k}\sin(kx) + C.
2
Apply the upper limit of integration x=π6x = \frac{\pi}{6}
\frac{1}{3}\sin\left(3 \cdot \frac{\pi}{6}\right) = \frac{1}{3}\sin\left(\frac{\pi}{2}\right) = \frac{1}{3}(1) = \frac{1}{3}
Substituting the upper limit into the antiderivative.
3
Apply the lower limit of integration x=0x = 0 and subtract
\frac{1}{3} - \frac{1}{3}\sin(0) = \frac{1}{3} - 0 = \frac{1}{3}
Evaluating the antiderivative at the limits according to the Fundamental Theorem of Calculus.

Anahtar Kavram

Definite Integration of Trigonometric Functions
Tahmini Süre:1m 30s
Soru 14Soru →

What is the value of the definite integral ∫13(3x2−2x) dx\int_{1}^{3} (3x^2 - 2x) \, dx?

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Cevap: 1818

Cevap

18
Integrating 3x2−2x3x^2 - 2x gives the antiderivative F(x)=x3−x2F(x) = x^3 - x^2. Evaluating F(3)−F(1)F(3) - F(1) gives (27−9)−(1−1)=18−0=18(27 - 9) - (1 - 1) = 18 - 0 = 18.

Adım Adım Çözüm

1
Find the indefinite integral of the function
\int (3x^2 - 2x) \, dx = x^3 - x^2
Apply the power rule of integration to each term: \int 3x^2 dx = x^3 and \int 2x dx = x^2.
2
Evaluate the antiderivative at the upper limit x = 3
(3)^3 - (3)^2 = 27 - 9 = 18
Substitute the upper limit into the antiderivative.
3
Evaluate the antiderivative at the lower limit x = 1
(1)^3 - (1)^2 = 1 - 1 = 0
Substitute the lower limit into the antiderivative.
4
Subtract the lower limit evaluation from the upper limit evaluation
18 - 0 = 18
By the Fundamental Theorem of Calculus, \int_{a}^{b} f(x) dx = F(b) - F(a).

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Fundamental Theorem of Calculus for Definite Polynomial Integrals
Tahmini Süre:1m 0s
Soru 15Soru →

Given that ∫1a(3x2−2x) dx=48\int_{1}^{a} (3x^2 - 2x) \, dx = 48, where a>1a > 1 is a constant, find the value of aa.

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Cevap: 4

Cevap

The value of the constant upper limit is 4.
Integrating 3x2−2x3x^2 - 2x yields x3−x2x^3 - x^2. Applying limits from 11 to aa gives (a3−a2)−(1−1)=a3−a2(a^3 - a^2) - (1 - 1) = a^3 - a^2. Setting a3−a2=48a^3 - a^2 = 48, solving for a>1a > 1 gives a=4a = 4 because 43−42=64−16=484^3 - 4^2 = 64 - 16 = 48.

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1
Integrate the polynomial function with respect to xx
∫(3x2−2x) dx=x3−x2+C\int (3x^2 - 2x) \, dx = x^3 - x^2 + C
Using the power rule of integration ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1} for each term.
2
Evaluate the antiderivative between the upper limit aa and lower limit 11
[x3−x2]1a=(a3−a2)−(13−12)=a3−a2[x^3 - x^2]_1^a = (a^3 - a^2) - (1^3 - 1^2) = a^3 - a^2
By the Fundamental Theorem of Calculus, ∫bcf(x)dx=F(c)−F(b)\int_{b}^{c} f(x)dx = F(c) - F(b).
3
Equate the expression to the given total integral value and solve for aa
a3−a2=48  ⟹  a=4a^3 - a^2 = 48 \implies a = 4
Substituting a=4a=4 yields 43−42=64−16=484^3 - 4^2 = 64 - 16 = 48, which satisfies the equation.

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Definite Integrals with Unknown Limits
Soru 16Soru →

Evaluate the definite integral ∫03(x2+2) dx\int_{0}^{3} (x^2 + 2) \, dx.

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Cevap: 15

Cevap

The value of the definite integral is 15.
Integrating x2+2x^2 + 2 with respect to xx yields [x33+2x]03\left[\frac{x^3}{3} + 2x\right]_{0}^{3}. Substituting the upper bound x=3x = 3 gives 273+6=15\frac{27}{3} + 6 = 15. Substituting the lower bound x=0x = 0 gives 00. The net value is 15−0=1515 - 0 = 15.

Adım Adım Çözüm

1
Find the antiderivative of x2+2x^2 + 2
x33+2x\frac{x^3}{3} + 2x
Apply the power rule of integration to each term.
2
Substitute the upper limit x=3x = 3
15
\frac{3^3}{3} + 2(3) = 9 + 6 = 15
3
Substitute the lower limit x=0x = 0
0
033+2(0)=0\frac{0^3}{3} + 2(0) = 0
4
Compute the difference between upper and lower limit values
15
15 - 0 = 15

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Definite Integration of Polynomial Functions
Tahmini Süre:45s
Soru 17Soru →

What is the value of the definite integral ∫04x dx\int_{0}^{4} \sqrt{x} \, dx?

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Cevap: 163\frac{16}{3}

Cevap

163\frac{16}{3}
Applying the power rule to x1/2x^{1/2} gives 23x3/2\frac{2}{3}x^{3/2}. Substituting the upper limit x=4x = 4 yields 23(4)3/2=23(8)=163\frac{2}{3}(4)^{3/2} = \frac{2}{3}(8) = \frac{16}{3}, and substituting the lower limit x=0x = 0 yields 0. Subtracting the lower limit result from the upper limit result gives 163\frac{16}{3}.

Adım Adım Çözüm

1
Express the integrand with a fractional exponent
x=x1/2\sqrt{x} = x^{1/2}
Rewriting the square root as a fractional exponent allows the application of the power rule of integration.
2
Find the antiderivative using the power rule of integration
∫x1/2 dx=x1/2+11/2+1=x3/23/2=23x3/2\int x^{1/2} \, dx = \frac{x^{1/2 + 1}}{1/2 + 1} = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}
The power rule states that ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1} for n≠−1n \neq -1.
3
Evaluate the antiderivative at the limits 4 and 0
[23x3/2]04=23(4)3/2−23(0)3/2=23(8)−0=163\left[ \frac{2}{3}x^{3/2} \right]_{0}^{4} = \frac{2}{3}(4)^{3/2} - \frac{2}{3}(0)^{3/2} = \frac{2}{3}(8) - 0 = \frac{16}{3}
By the Fundamental Theorem of Calculus, evaluate F(b)−F(a)F(b) - F(a).

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Definite Integration using the Power Rule
Soru 18Soru →

Find the total area of the region bounded by the curve y=3x2−6xy = 3x^2 - 6x, the xx-axis, and the vertical lines x=0x = 0 and x=3x = 3.

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Cevap: 8

Cevap

The total area bounded by the curve and the x-axis between x = 0 and x = 3 is 8 square units.
To find the total geometric area bounded by a curve and the x-axis, we must split the integral at any x-intercepts within the domain. For y=3x2−6xy = 3x^2 - 6x, the x-intercepts are x=0x = 0 and x=2x = 2. Between x=0x = 0 and x=2x = 2, the curve lies below the x-axis, giving an area magnitude of 44. Between x=2x = 2 and x=3x = 3, the curve lies above the x-axis, giving an area magnitude of 44. Summing these positive magnitudes gives a total area of 88.

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1
Find the roots of the curve y=3x2−6xy = 3x^2 - 6x within the given interval [0,3][0, 3].
Setting 3x2−6x=03x^2 - 6x = 0 yields 3x(x−2)=03x(x - 2) = 0, giving x=0x = 0 and x=2x = 2.
Roots inside the integration boundaries indicate where the curve crosses the x-axis, changing the sign of yy.
2
Determine the position of the curve relative to the x-axis on each sub-interval.
On [0,2][0, 2], y≤0y \le 0 (below the x-axis). On [2,3][2, 3], y≥0y \ge 0 (above the x-axis).
Geometric area must be non-negative, so regions below the x-axis require integrating −y-y or taking the absolute value of the integral.
3
Evaluate the area A1A_1 for the region below the x-axis from x=0x = 0 to x=2x = 2.
A1=∫02(6x−3x2) dx=[3x2−x3]02=(3(4)−8)−0=4A_1 = \int_{0}^{2} (6x - 3x^2) \, dx = \left[ 3x^2 - x^3 \right]_{0}^{2} = (3(4) - 8) - 0 = 4.
Integrating −y=6x−3x2-y = 6x - 3x^2 yields the positive magnitude of the area below the x-axis.
4
Evaluate the area A2A_2 for the region above the x-axis from x=2x = 2 to x=3x = 3.
A2=∫23(3x2−6x) dx=[x3−3x2]23=(33−3(32))−(23−3(22))=0−(−4)=4A_2 = \int_{2}^{3} (3x^2 - 6x) \, dx = \left[ x^3 - 3x^2 \right]_{2}^{3} = (3^3 - 3(3^2)) - (2^3 - 3(2^2)) = 0 - (-4) = 4.
Direct integration of yy on [2,3][2, 3] gives the area above the x-axis.
5
Combine the areas of both sub-regions.
Total Area =A1+A2=4+4=8= A_1 + A_2 = 4 + 4 = 8.
The total geometric area is the sum of the magnitudes of the areas of all separate bounded regions.

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Calculating area under curves crossing the x-axis by splitting definite integrals at real roots
Soru 19Soru →

What is the area of the region bounded by the curve y=6x−x2y = 6x - x^2 and the line y=2xy = 2x?

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Cevap: 323\frac{32}{3} square units

Cevap

323\frac{32}{3} square units
Equating y=6x−x2y = 6x - x^2 and y=2xy = 2x gives intersection points x=0x = 0 and x=4x = 4. Integrating the upper curve minus the lower line, ∫04(4x−x2) dx=[2x2−x33]04=32−643=323\int_{0}^{4} (4x - x^2) \, dx = \left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4} = 32 - \frac{64}{3} = \frac{32}{3} square units.

Adım Adım Çözüm

1
Find the points of intersection between the curve and the line
x=0x = 0 and x=4x = 4
Set 6x−x2=2x  ⟹  4x−x2=0  ⟹  x(4−x)=06x - x^2 = 2x \implies 4x - x^2 = 0 \implies x(4 - x) = 0 to find the integration bounds.
2
Set up the definite integral for the area between the two curves
A=∫04((6x−x2)−2x) dx=∫04(4x−x2) dxA = \int_{0}^{4} ((6x - x^2) - 2x) \, dx = \int_{0}^{4} (4x - x^2) \, dx
The area between two functions f(x)f(x) and g(x)g(x) from x=ax=a to x=bx=b is given by ∫ab(f(x)−g(x)) dx\int_{a}^{b} (f(x) - g(x)) \, dx where f(x)≥g(x)f(x) \ge g(x) on [a,b][a, b].
3
Compute the indefinite integral of the integrand
∫(4x−x2) dx=2x2−x33\int (4x - x^2) \, dx = 2x^2 - \frac{x^3}{3}
Apply the standard power rule of integration ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}.
4
Evaluate the definite integral using the limits 00 and 44
[2x2−x33]04=(2(4)2−433)−0=32−643=323\left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4} = \left( 2(4)^2 - \frac{4^3}{3} \right) - 0 = 32 - \frac{64}{3} = \frac{32}{3} square units
Substitute the upper limit x=4x = 4 and lower limit x=0x = 0 and simplify fractions.

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Area Between Two Curves
Tahmini Süre:2m 0s
Soru 20Soru →

What is the value of the definite integral ∫13(4x−1) dx\int_{1}^{3} (4x - 1) \, dx?

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Cevap: 14

Cevap

14
The antiderivative of 4x−14x - 1 is 2x2−x2x^2 - x. Evaluating at the upper boundary x=3x = 3 gives 2(3)2−3=152(3)^2 - 3 = 15, and at the lower boundary x=1x = 1 gives 2(1)2−1=12(1)^2 - 1 = 1. Subtracting the lower limit evaluation from the upper limit evaluation yields 15−1=1415 - 1 = 14.

Adım Adım Çözüm

1
Find the indefinite integral of the integrand 4x−14x - 1.
∫(4x−1) dx=2x2−x\int (4x - 1) \, dx = 2x^2 - x
Using the power rule of integration, ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}.
2
Substitute the upper limit x=3x = 3 into the antiderivative.
2(3)^2 - (3) = 2(9) - 3 = 18 - 3 = 15
Evaluating F(b)F(b) for F(x)=2x2−xF(x) = 2x^2 - x at b=3b = 3.
3
Substitute the lower limit x=1x = 1 into the antiderivative.
2(1)^2 - (1) = 2 - 1 = 1
Evaluating F(a)F(a) for F(x)=2x2−xF(x) = 2x^2 - x at a=1a = 1.
4
Subtract the lower limit result from the upper limit result.
15 - 1 = 14
By the Fundamental Theorem of Calculus, ∫abf(x) dx=F(b)−F(a)\int_{a}^{b} f(x) \, dx = F(b) - F(a).

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Evaluation of Definite Integrals using the Fundamental Theorem of Calculus
Tahmini Süre:1m 0s
Sayfa 1 / 2Sonraki
Definite Integrals and Area Under Curves Alıştırma Soruları — JAMB UTME | Examkin