Question

use geometry (not riemann sums) to evaluate the definite integral. sketch the graph of the integrand, show the region in question, and interpret your result.
int_{-4}^{3}(-2|x|)dx
choose the correct graph below.

Explanation:

Step1: Split integral at x=0

$\int_{-4}^{3} -2|x|dx = \int_{-4}^{0} -2|x|dx + \int_{0}^{3} -2|x|dx$

Step2: Simplify |x| for x<0

For $x<0$, $|x|=-x$, so $-2|x|=2x$. $\int_{-4}^{0} 2xdx$

Step3: Area of left triangle

Triangle with base=4, height=8. Area=$\frac{1}{2}*4*8=16$. Integral=-16 (below x-axis)

Step4: Simplify |x| for x≥0

For $x≥0$, $|x|=x$, so $-2|x|=-2x$. $\int_{0}^{3} -2xdx$

Step5: Area of right triangle

Triangle with base=3, height=6. Area=$\frac{1}{2}*3*6=9$. Integral=-9 (below x-axis)

Step6: Total integral

Sum: $-16 + (-9) = -25$

Answer:

-25