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Let $\int^4_{-2}f(x)dx = 3$, $\int^0_{-2}f(x)dx=10$, $\int^4_2f(x)dx=2$

Find $\int^2_0f(x)dx = -9$

and

$\int^0_2(3f(x) - 10)dx=??$

I figured out the first part is -9 since 10+2-9=3.

I thought the second part would be $3(-9)-10=-37$ since f(x) from 0 to 2 =-9 but that isn't correct. What am I doing wrong?

Jrow
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  • It's not from $0$ to $2$ it's from $2$ to $0$. You need another minus sign. –  Jan 08 '18 at 19:20

1 Answers1

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$\displaystyle\int_{2}^{0}(3f(x)-10)dx=-\int_{0}^{2}(3f(x)-10)dx=-3\int_{0}^{2}f(x)dx+10\int_{0}^{2}1dx=-3\int_{0}^{2}f(x)dx+20.$

user284331
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