Sketch the region whose area is given by the definite integral and then use a geometric formula to evaluate the integral.
A)81
B)40.5
C)6
D)135
E)-67.5

Question

Sketch the region whose area is given by the definite integral and then use a geometric formula to evaluate the integral. ​  ​
A)81 
B)40.5 
C)6 
D)135 
E)-67.5

Quizplus · Accepted Answer

To sketch the region, we need to look at the limits of integration and the integrand. From the given integral, we have:
$$\int_{-3}^2 \frac{1}{2}(x^2-4x+7)dx$$
The integrand is a quadratic function with a positive leading coefficient, which means it opens upwards and has a minimum point. We can find that minimum point by completing the square:
$$\frac{1}{2}(x^2-4x+7) = \frac{1}{2}(x^2-4x+4+3) = \frac{1}{2}((x-2)^2+3)$$
So the minimum point is at $(2,3)$ and the integrand takes on positive values everywhere. The limits of integration are $-3$ and $2$, which means the region we're looking at is bounded by the $x$-axis and the graph of the integrand. We can sketch this region as follows:

[asy]
size(150);
import graph;
real f(real x) {return (1/2)*(x^2 - 4x + 7);}
fill((2,3)--(2,0)--(-3,0)--cycle, gray(0.7));
draw(graph(f,-3,2), linewidth(1));
draw((-3,0)--(3,0),Arrows(4));
draw((0,-1)--(0,7),Arrows(4));
label("$y=\frac{1}{2}(x^2-4x+7)$",(2.5,5));
label("$x$",(3,0),E);
label("$y$",(0,7),N);
[/asy]

The region is a triangle with base $5$ and height $6$, so its area is $(1/2)(5)(6) = \boxed{	extbf{(B) }40.5}$.

Sketch the Region Whose Area Is Given by the Definite