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 ${\int}_{6}^{-3}h\left(t\right)dt$

a) Suppose that

b) Suppose that h is continuous and that

puhnut1m

Beginner2021-12-14Added 33 answers

Step 1

a)${\int}_{-9}^{-6}g\left(t\right)dt=6$

$\Rightarrow {\int}_{-9}^{-6}\frac{g\left(x\right)}{6}dx=11$

$\Rightarrow \frac{1}{6}{\int}_{-9}^{-6}g\left(x\right)dx=\frac{1}{6}{x}^{6}=1$

Step 2

$\Rightarrow {\int}_{-6}^{-9}-g\left(t\right)dt=?$

$\{\because {\int}_{a}^{b}f\left(x\right)dx=-{\int}_{b}^{a}f\left(x\right)dx\}$

$(-){\int}_{-6}^{-9}g\left(t\right)dt=(-)x(-1){\int}_{-9}^{-6}g\left(t\right)dt=(+){\int}_{-9}^{-6}g\left(t\right)dt=+6$

$\{{\int}_{-9}^{-6}\frac{g\left(x\right)d}{6}dx=1+{\int}_{-6}^{-9}-g\left(t\right)dt\}=+6$

a)

Step 2

Janet Young

Beginner2021-12-15Added 32 answers

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

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

$\{{\int}_{-3}^{6}h\left(t\right)dt=-8+{\int}_{6}^{-3}h\left(t\right)dt=+8\}$

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