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The base of a solid is the region enclosed by the graphs of 
y=x^(2)-5x+7 and 
y=3, between 
x=1 and 
x=4.
Cross sections of the solid perpendicular to the 
x-axis are rectangles whose height is 
x.
Which one of the definite integrals gives the volume of the solid?
Choose 1 answer:
(A) 
int_(1)^(4)(x^(2)-5x+4)*xdx
(B) 
piint_(1)^(4)(x^(2)-5x+4)^(2)dx
(C) 
int_(1)^(4)(x^(2)-5x+4)^(2)dx
(D) 
int_(1)^(4)(-x^(2)+5x-4)*xdx

The base of a solid is the region enclosed by the graphs of y=x25x+7 y=x^{2}-5 x+7 and y=3 y=3 , between x=1 x=1 and x=4 x=4 .\newlineCross sections of the solid perpendicular to the x x -axis are rectangles whose height is x x .\newlineWhich one of the definite integrals gives the volume of the solid?\newlineChoose 11 answer:\newline(A) 14(x25x+4)xdx \int_{1}^{4}\left(x^{2}-5 x+4\right) \cdot x d x \newline(B) π14(x25x+4)2dx \pi \int_{1}^{4}\left(x^{2}-5 x+4\right)^{2} d x \newline(C) 14(x25x+4)2dx \int_{1}^{4}\left(x^{2}-5 x+4\right)^{2} d x \newline(D) 14(x2+5x4)xdx \int_{1}^{4}\left(-x^{2}+5 x-4\right) \cdot x d x

Full solution

Q. The base of a solid is the region enclosed by the graphs of y=x25x+7 y=x^{2}-5 x+7 and y=3 y=3 , between x=1 x=1 and x=4 x=4 .\newlineCross sections of the solid perpendicular to the x x -axis are rectangles whose height is x x .\newlineWhich one of the definite integrals gives the volume of the solid?\newlineChoose 11 answer:\newline(A) 14(x25x+4)xdx \int_{1}^{4}\left(x^{2}-5 x+4\right) \cdot x d x \newline(B) π14(x25x+4)2dx \pi \int_{1}^{4}\left(x^{2}-5 x+4\right)^{2} d x \newline(C) 14(x25x+4)2dx \int_{1}^{4}\left(x^{2}-5 x+4\right)^{2} d x \newline(D) 14(x2+5x4)xdx \int_{1}^{4}\left(-x^{2}+5 x-4\right) \cdot x d x
  1. Define Volume Integral: To find the volume of the solid, we need to integrate the area of the cross-sections along the xx-axis from x=1x=1 to x=4x=4. The area of each rectangular cross-section is given by the difference in the yy-values of the two functions (y=3y=3 and y=x25x+7y=x^2-5x+7) times the height of the rectangle, which is xx. So, the area A(x)A(x) of a cross-section at a point xx is A(x)=(3(x25x+7))xA(x) = (3 - (x^2-5x+7)) \cdot x.
  2. Simplify Area Expression: First, simplify the expression for A(x)A(x) by distributing xx and combining like terms:\newlineA(x)=(3x2+5x7)xA(x) = (3 - x^2 + 5x - 7) \cdot x\newlineA(x)=(x2+5x4)xA(x) = (-x^2 + 5x - 4) \cdot x\newlineA(x)=x3+5x24xA(x) = -x^3 + 5x^2 - 4x
  3. Integrate to Find Volume: Now, we need to integrate A(x)A(x) from x=1x=1 to x=4x=4 to find the volume VV of the solid:\newlineV=x=1x=4A(x)dxV = \int_{x=1}^{x=4} A(x) \, dx\newlineV=x=1x=4(x3+5x24x)dxV = \int_{x=1}^{x=4} (-x^3 + 5x^2 - 4x) \, dx\newlineThis corresponds to one of the answer choices.
  4. Match with Answer Choices: Comparing the integral we found with the answer choices, we see that it matches with choice DD: \newlineDD extstylex=1x=4(x3+5x24x)dx extstyle \int_{x=1}^{x=4} (-x^3 + 5x^2 - 4x) \, dx

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