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Zorluk: ZorDefinite Integrals and Area Under Curves

The area of the region bounded by the curve y=3x2−4x+3y = 3x^2 - 4x + 3, the xx-axis, and the vertical lines x=0x = 0 and x=kx = k (where k>0k > 0) is 1818 square units. What is the value of kk?

Cevap: 3

Cevap

The value of kk is 33.
The area under the non-negative curve y=3x2−4x+3y = 3x^2 - 4x + 3 from x=0x = 0 to x=kx = k is found by calculating ∫0k(3x2−4x+3) dx=k3−2k2+3k\int_{0}^{k} (3x^2 - 4x + 3) \, dx = k^3 - 2k^2 + 3k. Setting this expression equal to 1818 gives k3−2k2+3k−18=0k^3 - 2k^2 + 3k - 18 = 0. By the factor theorem, testing k=3k = 3 yields 33−2(3)2+3(3)−18=03^3 - 2(3)^2 + 3(3) - 18 = 0. Factoring as (k−3)(k2+k+6)=0(k - 3)(k^2 + k + 6) = 0 confirms k=3k = 3 as the only real solution.

Adım Adım Çözüm

1
Set up the definite integral for the area bounded by the curve and the x-axis
\int_{0}^{k} (3x^2 - 4x + 3) \, dx = 18
The curve y=3x2−4x+3y = 3x^2 - 4x + 3 lies entirely above the x-axis for all real xx because its leading coefficient is positive and its discriminant (−4)2−4(3)(3)=−20<0(-4)^2 - 4(3)(3) = -20 < 0.
2
Evaluate the definite integral in terms of kk
\left[ x^3 - 2x^2 + 3x \right]_0^k = (k^3 - 2k^2 + 3k) - 0 = k^3 - 2k^2 + 3k
Applying the fundamental theorem of calculus by integrating term by term.
3
Form and simplify the polynomial equation
k^3 - 2k^2 + 3k - 18 = 0
Equating the definite integral expression to the given area value of 18.
4
Solve for real values of k>0k > 0
k = 3
Using the factor theorem on k3−2k2+3k−18=0k^3 - 2k^2 + 3k - 18 = 0, k=3k = 3 yields zero (27−18+9−18=027 - 18 + 9 - 18 = 0). Factoring out (k−3)(k - 3) gives (k−3)(k2+k+6)=0(k - 3)(k^2 + k + 6) = 0, where k2+k+6=0k^2 + k + 6 = 0 has complex roots.

Anahtar Kavram

Determining an unknown boundary limit of a definite integral representing area under a curve
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Konular

Definite Integrals and Area Under Curves

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