Osvaldo Apodaca

2021-12-13

Solve the problem.
a) Suppose that ${\int }_{-9}^{-6}g\left(t\right)dt=6$. Find ${\int }_{-9}^{-6}\frac{g\left(x\right)}{6}dx$ and ${\int }_{-6}^{-9}-g\left(t\right)dt$
b) Suppose that h is continuous and that ${\int }_{-3}^{3}h\left(x\right)dx=4$ and ${\int }_{3}^{6}h\left(x\right)dx=-12$. Find ${\int }_{-3}^{6}h\left(t\right)dt$ and ${\int }_{6}^{-3}h\left(t\right)dt$

puhnut1m

Step 1
a) ${\int }_{-9}^{-6}g\left(t\right)dt=6$
$⇒{\int }_{-9}^{-6}\frac{g\left(x\right)}{6}dx=11$
$⇒\frac{1}{6}{\int }_{-9}^{-6}g\left(x\right)dx=\frac{1}{6}{x}^{6}=1$
Step 2
$⇒{\int }_{-6}^{-9}-g\left(t\right)dt=?$
$\left\{\because {\int }_{a}^{b}f\left(x\right)dx=-{\int }_{b}^{a}f\left(x\right)dx\right\}$
$\left(-\right){\int }_{-6}^{-9}g\left(t\right)dt=\left(-\right)x\left(-1\right){\int }_{-9}^{-6}g\left(t\right)dt=\left(+\right){\int }_{-9}^{-6}g\left(t\right)dt=+6$
$\left\{{\int }_{-9}^{-6}\frac{g\left(x\right)d}{6}dx=1+{\int }_{-6}^{-9}-g\left(t\right)dt\right\}=+6$

Janet Young

b) ${\int }_{-3}^{3}h\left(x\right)dx=4$, ${\int }_{3}^{6}h\left(x\right)dx=-12$, ${\int }_{-3}^{6}h\left(t\right)dt$
$\because {\int }_{-3}^{6}h\left(t\right)dt={\int }_{-3}^{3}h\left(x\right)dx+{\int }_{3}^{6}h\left(x\right)dx$
$\left\{{\int }_{-3}^{6}h\left(t\right)dt=-8+{\int }_{6}^{-3}h\left(t\right)dt=+8\right\}$

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