Graph the integrand and use areas to evaluate the integral.
-$$\int _ { - 10 } ^ { 5 } ( - 2 x + 10 ) d x$$
A) 450
B) 300
C) 75
D) 225

Question

Quizplus · Accepted Answer

The integrand is a linear function with negative slope, starting at $y=10$ and decreasing to $y=-20$. 
To find the value of the integral, we need to calculate the area of the region between the $x$-axis and the function. 
Divide the region into a rectangle and a triangle, as shown:

[asy]
unitsize(0.15cm);
import graph;
defaultpen(linewidth(.7));
real f(real x)
{
return -2*x+10;
}

fill((-10,0)--(-10,10)--(5,10)--(5,-20)--cycle,gray(0.7));
draw(graph(f,-10,5),linewidth(1.2));
draw((-10,0)--(5,0),Arrows(4));
draw((0,-20)--(0,10),Arrows(4));
label("$y=-2x+10$",(5.2,-20));
[/asy]

The rectangle has base $b = 15$ (from $x=-10$ to $x=5$) and height $h = 10$, so its area is $A_r = bh = 15 \cdot 10 = 150$. 
The triangle has base $b = 5-(-10) = 15$ and height $h = |-20-10| = 30$, so its area is $A_t = \frac{1}{2}bh = \frac{1}{2} \cdot 15 \cdot 30 = 225$. 
The total area is $A = A_r + A_t = 150 + 225 = 375$. 
Since the function is below the $x$-axis for $x<2.5$, we need to subtract the area of the region between $x=-10$ and $x=2.5$:

[asy]
unitsize(0.15cm);
import graph;
defaultpen(linewidth(.7));
real f(real x)
{
return -2*x+10;
}

fill((-10,0)--(-10,10)--(2.5,10)--cycle,gray(0.7));
draw(graph(f,-10,5),linewidth(1.2));
draw((-10,0)--(5,0),Arrows(4));
draw((0,-20)--(0,10),Arrows(4));
label("$y=-2x+10$",(5.2,-20));
[/asy]

This region is a rectangle with base $b = 2.5-(-10) = 12.5$ and height $h=10$, so its area is $A_1 = bh = 12.5 \cdot 10 = 125$. 
Therefore, the value of the integral is $-A_1+A = -125+375 = \boxed{	extbf{(D) }225}$.

Graph the Integrand and Use Areas to Evaluate the Integral ∫−105(−2x+10)dx\int _ { - 10 } ^ { 5 } ( - 2 x + 10 ) d x∫−105​(−2x+10)dx

Graph the Integrand and Use Areas to Evaluate the Integral $\int _ { - 10 } ^ { 5 } ( - 2 x + 10 ) d x$