Calculus Alıştırma Soruları — JAMB UTME — Sayfa 9

Calculate the gradient of the normal line to the curve $y = \frac{6}{x}$ at the point where $x = 3$ .

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Cevap: 1.5

Cevap

The gradient of the normal line is 1.5.

Differentiating

y = 6x^{-1}

yields

\frac{dy}{dx} = -\frac{6}{x^2}

. Evaluating this derivative at

x = 3

gives the tangent gradient

m_t = -\frac{6}{9} = -\frac{2}{3}

. Because the normal line is perpendicular to the tangent line, its gradient is the negative reciprocal

m_n = -\frac{1}{m_t} = \frac{3}{2} = 1.5

.

Adım Adım Çözüm

1

Differentiate the function

y = 6x^{-1}

with respect to

x

\frac{dy}{dx} = -6x^{-2} = -\frac{6}{x^2}

The first derivative represents the formula for the tangent gradient to the curve at any given point.

2

Evaluate the derivative at

x = 3

to find the tangent slope (

m_t

)

m_t = -\frac{6}{3^2} = -\frac{6}{9} = -\frac{2}{3}

Substituting the given x-coordinate into the derivative gives the exact slope of the tangent at that point.

3

Calculate the normal slope (

m_n

) as the negative reciprocal of

m_t

m_n = -\frac{1}{m_t} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} = 1.5

The normal line is perpendicular to the tangent line, meaning

m_t \cdot m_n = -1

.

Anahtar Kavram

The gradient of the normal to a curve

y = f(x)

at

x = a

is the negative reciprocal of the derivative evaluated at that point:

m_n = -\frac{1}{f'(a)}

.

Soru 162Soru →

What is the value of the definite integral $\int_{1}^{2} \left(3x^2 - \frac{4}{x^2}\right) \, dx$ ?

5

6

9

10

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Cevap: 5

Cevap

5

Integrating

3x^2 - 4x^{-2}

gives

x^3 + \frac{4}{x}

. Substituting the upper bound

x=2

gives

8 + 2 = 10

, and substituting the lower bound

x=1

gives

1 + 4 = 5

. The difference is

10 - 5 = 5

.

Adım Adım Çözüm

1

Rewrite the integrand with a negative exponent

3x^2 - 4x^{-2}

Converting reciprocal powers into negative exponents simplifies applying the power rule of integration.

2

Find the indefinite integral (antiderivative)

\int (3x^2 - 4x^{-2}) \, dx = \frac{3x^3}{3} - \frac{4x^{-1}}{-1} = x^3 + \frac{4}{x}

Apply the power rule

\int x^n \, dx = \frac{x^{n+1}}{n+1}

to each term.

3

Evaluate the antiderivative at the upper limit

x = 2

2^3 + \frac{4}{2} = 8 + 2 = 10

Substitute the upper limit into the antiderivative expression.

4

Evaluate the antiderivative at the lower limit

x = 1

1^3 + \frac{4}{1} = 1 + 4 = 5

Substitute the lower limit into the antiderivative expression.

5

Subtract the lower limit value from the upper limit value

10 - 5 = 5

According to the Fundamental Theorem of Calculus,

\int_{a}^{b} f(x)\,dx = F(b) - F(a)

.

Anahtar Kavram

Definite Integration of Polynomial and Power Functions

Soru 163Soru →

If $\int (8x^3 - 9\sin(3x) + 2) \, dx = ax^4 + b\cos(3x) + cx + C$ , where $a$ , $b$ , and $c$ are constant coefficients and $C$ is the constant of integration, what is the value of $a + b + c$ ?

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Cevap: 7

Cevap

The value of

a + b + c

is 7.

Integrating term-by-term yields

\int 8x^3 dx = 2x^4

,

\int -9\sin(3x) dx = 3\cos(3x)

, and

\int 2 dx = 2x

. Equating coefficients with

ax^4 + b\cos(3x) + cx

gives

a = 2

,

b = 3

, and

c = 2

. Therefore,

a + b + c = 2 + 3 + 2 = 7

.

Adım Adım Çözüm

1

Integrate the polynomial term

8x^3

\int 8x^3 \, dx = 2x^4

, identifying

a = 2

Apply the power rule of integration:

\int x^n \, dx = \frac{x^{n+1}}{n+1}

.

2

Integrate the trigonometric term

-9\sin(3x)

\int -9\sin(3x) \, dx = 3\cos(3x)

, identifying

b = 3

Apply the standard trigonometric integral formula:

\int \sin(kx) \, dx = -\frac{1}{k}\cos(kx)

.

3

Integrate the constant term

2

\int 2 \, dx = 2x

, identifying

c = 2

The integral of a constant

k

with respect to

x

is

kx

.

4

Calculate the requested sum

a + b + c

a + b + c = 2 + 3 + 2 = 7

Summing the coefficients derived from each term's antiderivative.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 164Soru →

If $y = \frac{e^{3x}}{\cos x + \sin x}$ , what is the value of $\frac{dy}{dx}$ at $x = 0$ ?

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Cevap: 2

Cevap

The value of

\frac{dy}{dx}

at

x = 0

is

2

.

Applying the quotient rule

\frac{dy}{dx} = \frac{v u' - u v'}{v^2}

to

y = \frac{e^{3x}}{\cos x + \sin x}

yields

\frac{dy}{dx} = \frac{3e^{3x}(\cos x + \sin x) - e^{3x}(\cos x - \sin x)}{(\cos x + \sin x)^2}

. Evaluating this expression at

x = 0

gives

\frac{3(1)(1) - (1)(1)}{1^2} = 2

.

Adım Adım Çözüm

1

Identify numerator

u(x)

and denominator

v(x)

for the quotient rule.

u(x) = e^{3x}

and

v(x) = \cos x + \sin x

.

The function

y

is expressed as a quotient of exponential and trigonometric functions.

2

Find the first derivatives of

u(x)

and

v(x)

.

u'(x) = 3e^{3x}

and

v'(x) = -\sin x + \cos x

.

Derivative of

e^{kx}

is

k e^{kx}

, derivative of

\cos x

is

-\sin x

, and derivative of

\sin x

is

\cos x

.

3

Substitute

u(x)

,

v(x)

,

u'(x)

, and

v'(x)

into the quotient rule formula

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

.

\frac{dy}{dx} = \frac{3e^{3x}(\cos x + \sin x) - e^{3x}(\cos x - \sin x)}{(\cos x + \sin x)^2}

.

The quotient rule is required to differentiate

\frac{u(x)}{v(x)}

.

4

Evaluate the derivative expression at

x = 0

.

\frac{dy}{dx}\Big|_{x=0} = \frac{3(1)(1 + 0) - 1(1 - 0)}{(1 + 0)^2} = \frac{3 - 1}{1} = 2

.

At

x = 0

,

e^0 = 1

,

\cos 0 = 1

, and

\sin 0 = 0

.

Anahtar Kavram

Differentiation of exponential and trigonometric functions using the quotient rule

Soru 165Soru →

If $y = e^{-2x} \sin(3x)$ , find $\frac{dy}{dx}$ .

e^{-2x} (3\cos(3x) + 2\sin(3x))

e^{-2x} (\cos(3x) - 2\sin(3x))

e^{-2x} (3\cos(3x) - 2\sin(3x))

-6e^{-2x} \cos(3x)

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Cevap: $e^{-2x} (3\cos(3x) - 2\sin(3x))$

Cevap

\frac{dy}{dx} = e^{-2x} (3\cos(3x) - 2\sin(3x))

Applying the product rule

\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}

to

u = e^{-2x}

and

v = \sin(3x)

yields

\frac{du}{dx} = -2e^{-2x}

and

\frac{dv}{dx} = 3\cos(3x)

. Substituting these terms gives

e^{-2x}(3\cos(3x)) + \sin(3x)(-2e^{-2x}) = e^{-2x}(3\cos(3x) - 2\sin(3x))

.

Adım Adım Çözüm

1

Identify the component functions for the product rule

Let

u = e^{-2x}

and

v = \sin(3x)

.

The function

y

is a product of an exponential function and a trigonometric function.

2

Differentiate each component using the chain rule

\frac{du}{dx} = -2e^{-2x}

and

\frac{dv}{dx} = 3\cos(3x)

.

\frac{d}{dx}(e^{kx}) = k e^{kx}

and

\frac{d}{dx}(\sin(kx)) = k \cos(kx)

where

k

is a constant.

3

Apply the product rule formula

\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

and factor out

e^{-2x}

\frac{dy}{dx} = e^{-2x}(3\cos(3x)) + \sin(3x)(-2e^{-2x}) = e^{-2x}(3\cos(3x) - 2\sin(3x))

.

Combining terms correctly gives the exact derivative.

Anahtar Kavram

Differentiation of Product of Transcendental Functions (Exponential and Trigonometric)

Soru 166Soru →

What is the $x$ -coordinate of the maximum stationary point of the curve $y = \sin(2x) - x$ in the interval $0 \le x \le \pi$ ?

\frac{\pi}{6}

\frac{5\pi}{6}

\frac{\pi}{3}

\pi

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Cevap: $\frac{\pi}{6}$

Cevap

The

x

-coordinate of the maximum stationary point is

\frac{\pi}{6}

.

Differentiating

y = \sin(2x) - x

gives

\frac{dy}{dx} = 2\cos(2x) - 1

. Setting this derivative to zero yields

\cos(2x) = \frac{1}{2}

, giving

x = \frac{\pi}{6}

and

x = \frac{5\pi}{6}

in the given interval. Checking the second derivative

\frac{d^2y}{dx^2} = -4\sin(2x)

at

x = \frac{\pi}{6}

gives

-2\sqrt{3} < 0

, which confirms that the maximum stationary point occurs at

\frac{\pi}{6}

.

Adım Adım Çözüm

1

Find the first derivative of

y = \sin(2x) - x

using the chain rule.

\frac{dy}{dx} = 2\cos(2x) - 1

Stationary points occur where the rate of change

\frac{dy}{dx} = 0

.

2

Set

\frac{dy}{dx} = 0

and solve for

x

within

0 \le x \le \pi

.

2\cos(2x) - 1 = 0 \implies \cos(2x) = \frac{1}{2} \implies 2x = \frac{\pi}{3}

or

2x = \frac{5\pi}{3} \implies x = \frac{\pi}{6}

or

x = \frac{5\pi}{6}

The trigonometric equation

\cos(\theta) = \frac{1}{2}

has solutions

\frac{\pi}{3}

and

\frac{5\pi}{3}

in

[0, 2\pi]

.

3

Evaluate the second derivative

\frac{d^2y}{dx^2}

to determine the nature of the stationary points.

\frac{d^2y}{dx^2} = -4\sin(2x)

. At

x = \frac{\pi}{6}

,

\frac{d^2y}{dx^2} = -4\sin\left(\frac{\pi}{3}\right) = -2\sqrt{3} < 0

.

A negative second derivative (

\frac{d^2y}{dx^2} < 0

) indicates a local maximum.

Anahtar Kavram

Determining maximum stationary points of trigonometric functions using the first and second derivative tests.

Soru 167Soru →

If $y = (2x^2 + 1)^3$ , what is the numerical value of $\frac{dy}{dx}$ evaluated at $x = 1$ ?

27

54

108

36

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Cevap: $108$

Cevap

108

Differentiating

y = (2x^2 + 1)^3

using the chain rule yields

\frac{dy}{dx} = 3(2x^2 + 1)^2 \cdot 4x = 12x(2x^2 + 1)^2

. Substituting

x = 1

gives

12(1)(3)^2 = 108

.

Adım Adım Çözüm

1

Identify inner and outer functions for the composite expression

Let

u = 2x^2 + 1

, so

y = u^3

.

The function requires the application of the Chain Rule.

2

Compute the derivatives of the outer and inner functions

\frac{dy}{du} = 3u^2 = 3(2x^2 + 1)^2

and

\frac{du}{dx} = 4x

.

Apply the power rule to both components.

3

Apply the Chain Rule formula

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

\frac{dy}{dx} = 3(2x^2 + 1)^2 \cdot (4x) = 12x(2x^2 + 1)^2

.

Multiply the outer derivative by the inner derivative.

4

Substitute

x = 1

into the derivative

\frac{dy}{dx}\Big|_{x=1} = 12(1)(2(1)^2 + 1)^2 = 12(3)^2 = 12 \cdot 9 = 108

.

Evaluate at the given x-value.

Anahtar Kavram

Chain Rule of Differentiation

Tahmini Süre:1m 0s

Soru 168Soru →

If $y = \ln(2 + \sin(3x)) + e^{4x}$ , find the value of $\frac{dy}{dx}$ at $x = 0$ .

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Cevap: 5.5

Cevap

The value of

\frac{dy}{dx}

at

x = 0

is

5.5

.

Differentiating

\ln(2 + \sin(3x))

by the chain rule gives

\frac{3\cos(3x)}{2 + \sin(3x)}

, and differentiating

e^{4x}

gives

4e^{4x}

. Evaluating

\frac{3\cos(3x)}{2 + \sin(3x)} + 4e^{4x}

at

x = 0

yields

\frac{3(1)}{2 + 0} + 4(1) = 1.5 + 4 = 5.5

.

Adım Adım Çözüm

1

Differentiate the logarithmic component

\ln(2 + \sin(3x))

using the chain rule

\frac{d}{dx}[\ln(2 + \sin(3x))] = \frac{3\cos(3x)}{2 + \sin(3x)}

By the chain rule,

\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx}

, where

u = 2 + \sin(3x)

and

\frac{du}{dx} = 3\cos(3x)

.

2

Differentiate the exponential component

e^{4x}

\frac{d}{dx}[e^{4x}] = 4e^{4x}

The standard rule for exponential differentiation states that

\frac{d}{dx}[e^{kx}] = k e^{kx}

.

3

Combine the results to state the derivative function

\frac{dy}{dx}

\frac{dy}{dx} = \frac{3\cos(3x)}{2 + \sin(3x)} + 4e^{4x}

The derivative of a sum of functions is the sum of their individual derivatives.

4

Evaluate

\frac{dy}{dx}

at

x = 0

\frac{dy}{dx}\Big|_{x=0} = \frac{3\cos(0)}{2 + \sin(0)} + 4e^0 = \frac{3(1)}{2 + 0} + 4(1) = 1.5 + 4 = 5.5

Substituting

x = 0

uses the values

\cos(0) = 1

,

\sin(0) = 0

, and

e^0 = 1

.

Anahtar Kavram

Differentiation of Logarithmic, Trigonometric, and Exponential Functions using the Chain Rule

Soru 169Soru →

The rate of change of a function $f(x)$ with respect to $x$ is defined by $f'(x) = 3x^2 - 4x + 6\sin(3x)$ . If $f(0) = 7$ , determine the value of the constant of integration, $C$ .

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Cevap: 9

Cevap

The constant of integration

C

is equal to 9.

Integrating

f'(x) = 3x^2 - 4x + 6\sin(3x)

gives

f(x) = x^3 - 2x^2 - 2\cos(3x) + C

. Substituting

x = 0

and

f(0) = 7

leads to

7 = 0 - 0 - 2(1) + C

, which simplifies to

C = 9

.

Adım Adım Çözüm

1

Integrate the rate of change function

f'(x) = 3x^2 - 4x + 6\sin(3x)

with respect to

x

.

f(x) = x^3 - 2x^2 - 2\cos(3x) + C

Using the power rule

\int x^n dx = \frac{x^{n+1}}{n+1}

and trigonometric integration rule

\int \sin(kx) dx = -\frac{1}{k}\cos(kx)

.

2

Substitute the initial condition

x = 0

and

f(0) = 7

into the expression for

f(x)

.

7 = (0)^3 - 2(0)^2 - 2\cos(3 \cdot 0) + C

The curve passes through

x = 0

with value

y = 7

.

3

Evaluate the trigonometric term at zero and solve for

C

.

7 = -2(1) + C \implies C = 9

Since

\cos(0) = 1

, the expression simplifies to

7 = -2 + C

, yielding

C = 9

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions with Boundary Conditions

Soru 170Soru →

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

5

2

-1

-5

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Cevap: $5$

Cevap

The value of the definite integral is

5

.

The antiderivative of

3\sin x + 2\cos x

is

-3\cos x + 2\sin x

. Evaluating this expression at the upper limit

x = \frac{\pi}{2}

gives

2

, and at the lower limit

x = 0

gives

-3

. Applying the Fundamental Theorem of Calculus gives

2 - (-3) = 5

.

Adım Adım Çözüm

1

Find the indefinite integral of

3\sin x + 2\cos x

\int (3\sin x + 2\cos x) \, dx = -3\cos x + 2\sin x

The antiderivative of

\sin x

is

-\cos x

and the antiderivative of

\cos x

is

\sin x

.

2

Evaluate the antiderivative at the upper limit

x = \frac{\pi}{2}

-3\cos\left(\frac{\pi}{2}\right) + 2\sin\left(\frac{\pi}{2}\right) = -3(0) + 2(1) = 2

Substitute

x = \frac{\pi}{2}

into the antiderivative expression.

3

Evaluate the antiderivative at the lower limit

x = 0

-3\cos(0) + 2\sin(0) = -3(1) + 2(0) = -3

Substitute

x = 0

into the antiderivative expression.

4

Subtract the lower limit value from the upper limit value

2 - (-3) = 2 + 3 = 5

By the Fundamental Theorem of Calculus,

\int_{a}^{b} f(x)\,dx = F(b) - F(a)

.

Anahtar Kavram

Definite Integration of Trigonometric Functions

Tahmini Süre:1m 30s

Soru 171Soru →

Given the function $f(x) = 5x^2 - 4x + 3$ , what is the numerical value of its derivative at $x = 2$ when evaluated using the first principles limit definition $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ ?

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Cevap: 16

Cevap

The numerical value of the derivative of

f(x) = 5x^2 - 4x + 3

at

x = 2

is 16.

Evaluating the first principles definition

\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

for

f(x) = 5x^2 - 4x + 3

yields the derivative

f'(x) = 10x - 4

. Substituting

x = 2

yields

10(2) - 4 = 16

.

Adım Adım Çözüm

1

Determine the expanded form of

f(x+h)

f(x+h) = 5(x+h)^2 - 4(x+h) + 3 = 5x^2 + 10xh + 5h^2 - 4x - 4h + 3

Evaluating the function at

x+h

requires expanding the square and distributing the constant factors.

2

Compute the difference

f(x+h) - f(x)

f(x+h) - f(x) = (5x^2 + 10xh + 5h^2 - 4x - 4h + 3) - (5x^2 - 4x + 3) = 10xh + 5h^2 - 4h

Subtracting

f(x)

cancels terms independent of

h

.

3

Form and simplify the difference quotient

\frac{f(x+h) - f(x)}{h}

\frac{10xh + 5h^2 - 4h}{h} = 10x + 5h - 4

Factoring

h

out of the numerator allows division by

h

for non-zero

h

.

4

Evaluate the limit as

h \to 0

f'(x) = \lim_{h \to 0} (10x + 5h - 4) = 10x - 4

Taking the limit produces the general derivative function

\frac{\mathrm{d}y}{\mathrm{d}x}

.

5

Substitute

x = 2

into the derivative

f'(2) = 10(2) - 4 = 16

Evaluating at

x = 2

gives the instantaneous rate of change at that point.

Anahtar Kavram

Differentiation from First Principles

Tahmini Süre:1m 30s

Soru 172Soru →

Given that $\int \left( k x^3 + 12\cos(3x) \right) dx = 4x^4 + 4\sin(3x) + C$ , where $C$ is the arbitrary constant of integration, what is the numerical value of the constant $k$ ?

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

The numerical value of the constant

k

is 16.

Integrating

kx^3 + 12\cos(3x)

with respect to

x

yields

\frac{k}{4}x^4 + 4\sin(3x) + C

. Comparing the coefficient of

x^4

with the given result

4x^4 + 4\sin(3x) + C

gives

\frac{k}{4} = 4

, which leads to

k = 16

.

Adım Adım Çözüm

1

Integrate the polynomial and trigonometric terms separately using standard integration rules.

\int (kx^3 + 12\cos(3x)) dx = \frac{k}{4}x^4 + 4\sin(3x) + C

Applying the power rule

\int x^n dx = \frac{x^{n+1}}{n+1}

gives

\int kx^3 dx = \frac{k}{4}x^4

, and applying

\int \cos(ax) dx = \frac{\sin(ax)}{a}

gives

\int 12\cos(3x) dx = \frac{12}{3}\sin(3x) = 4\sin(3x)

.

2

Equate the integrated expression to the right-hand side of the given equation.

\frac{k}{4}x^4 + 4\sin(3x) + C = 4x^4 + 4\sin(3x) + C

Both sides represent the same antiderivative of the function.

3

Equate corresponding coefficients of

x^4

to solve for

k

.

k = 16

\frac{k}{4} = 4 \implies k = 4 \times 4 = 16.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 173Soru →

What is the maximum value of the curve $y = \frac{x}{x^2 + 4}$ ?

2

\frac{1}{4}

-\frac{1}{4}

\frac{1}{2}

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{4}$

Cevap

The maximum value of the curve is

\frac{1}{4}

.

Differentiating

y = \frac{x}{x^2 + 4}

using the quotient rule gives

\frac{dy}{dx} = \frac{4 - x^2}{(x^2 + 4)^2}

. Setting

\frac{dy}{dx} = 0

yields critical points at

x = \pm 2

. Substituting

x = 2

into the original function gives

y = \frac{2}{8} = \frac{1}{4}

, which is the maximum value of the function.

Adım Adım Çözüm

1

Differentiate

y = \frac{x}{x^2 + 4}

with respect to

x

using the quotient rule.

\frac{dy}{dx} = \frac{(x^2 + 4)(1) - x(2x)}{(x^2 + 4)^2} = \frac{4 - x^2}{(x^2 + 4)^2}

Stationary points occur where the first derivative equals zero.

2

Set

\frac{dy}{dx} = 0

to solve for the stationary points.

4 - x^2 = 0 \implies x^2 = 4 \implies x = 2 \text{ or } x = -2

A rational expression equals zero when its numerator is zero.

3

Evaluate

y

at each critical point to determine the function values.

For

x = 2

:

y = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4}

. For

x = -2

:

y = \frac{-2}{(-2)^2 + 4} = -\frac{1}{4}

.

The question asks for the maximum value of

y

on the curve.

4

Compare the stationary values to select the maximum.

The maximum value is

\frac{1}{4}

at

x = 2

.

Since

\frac{1}{4} > -\frac{1}{4}

,

x = 2

corresponds to the maximum point.

Anahtar Kavram

Stationary Points and Maxima/Minima of Rational Functions

Tahmini Süre:1m 30s

Soru 174Soru →

Evaluate the algebraic limit:

\lim_{x \to 2} \frac{x^3 - 8}{x^2 + x - 6}

What is the value of this limit?

\frac{4}{5}

\frac{8}{5}

\frac{12}{5}

0

Cevabı ve açıklamayı göster

Cevap: $\frac{12}{5}$

Cevap

The correct value of the limit is

\frac{12}{5}

.

Evaluating the limit by direct substitution gives the indeterminate form

\frac{0}{0}

. Factoring the numerator

x^3 - 8 = (x - 2)(x^2 + 2x + 4)

and denominator

x^2 + x - 6 = (x - 2)(x + 3)

allows cancellation of

(x - 2)

. Evaluating

\frac{x^2 + 2x + 4}{x + 3}

at

x = 2

yields

\frac{12}{5}

.

Adım Adım Çözüm

1

Check for direct substitution

Substituting

x = 2

gives

\frac{2^3 - 8}{2^2 + 2 - 6} = \frac{0}{0}

, which is an indeterminate form.

Direct substitution results in

\frac{0}{0}

, requiring algebraic factorization.

2

Factor the numerator and the denominator

Numerator:

x^3 - 8 = (x - 2)(x^2 + 2x + 4)

Denominator:

x^2 + x - 6 = (x - 2)(x + 3)

Use the difference of cubes formula

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

and quadratic factorization.

3

Cancel the common factor and compute the limit

\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 3)} = \lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 3} = \frac{2^2 + 2(2) + 4}{2 + 3} = \frac{12}{5}

Eliminating the factor

(x - 2)

removes the removable discontinuity at

x = 2

.

Anahtar Kavram

Limits of Indeterminate Forms (0/0) using Factorization

Soru 175Soru →

What is the area of the region bounded by the curve $y = x^2 - 4$ , the $x$ -axis, and the lines $x = 0$ and $x = 3$ ?

\frac{23}{3}

square units

3

square units

\frac{16}{3}

square units

\frac{25}{3}

square units

Cevabı ve açıklamayı göster

Cevap: $\frac{23}{3}$ square units

Cevap

\frac{23}{3}

square units

The curve

y = x^2 - 4

intersects the x-axis at

x = 2

. To find the total enclosed area between

x = 0

and

x = 3

, the integral must be split into two parts:

[0, 2]

, where the curve is below the x-axis (yielding an area of

\frac{16}{3}

), and

[2, 3]

, where the curve is above the x-axis (yielding an area of

\frac{7}{3}

). Summing these absolute values gives

\frac{16}{3} + \frac{7}{3} = \frac{23}{3}

square units.

Adım Adım Çözüm

1

Find the x-intercept of the curve

y = x^2 - 4

within the interval

[0, 3]

.

x^2 - 4 = 0 \implies x = 2

. The curve lies below the x-axis for

0 \le x < 2

and above the x-axis for

2 < x \le 3

.

Total geometric area requires evaluating regions below and above the x-axis separately so negative integral values do not cancel positive area.

2

Calculate the area

A_1

of the region below the x-axis from

x = 0

to

x = 2

.

A_1 = \left| \int_{0}^{2} (x^2 - 4) \, dx \right| = \left| \left[ \frac{x^3}{3} - 4x \right]_{0}^{2} \right| = \left| \frac{8}{3} - 8 \right| = \left| -\frac{16}{3} \right| = \frac{16}{3}

square units.

The curve is below the x-axis, so taking the absolute value gives the true physical area.

3

Calculate the area

A_2

of the region above the x-axis from

x = 2

to

x = 3

.

A_2 = \int_{2}^{3} (x^2 - 4) \, dx = \left[ \frac{x^3}{3} - 4x \right]_{2}^{3} = (9 - 12) - \left( \frac{8}{3} - 8 \right) = -3 - \left(-\frac{16}{3}\right) = \frac{7}{3}

square units.

The curve lies above the x-axis on this interval, yielding a positive definite integral.

4

Sum the areas of the two regions to find the total bounded area.

\text{Total Area} = A_1 + A_2 = \frac{16}{3} + \frac{7}{3} = \frac{23}{3}

square units.

Adding the individual positive areas yields the total bounded area.

Anahtar Kavram

Area bounded by a curve that crosses the x-axis

Calculus

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