Question

# $\underset{a+c}{\overset{b+c}{\int }}f\left(x\right)dx$ is equal to (a) $\underset{a}{\overset{b}{\int }}f\left(x-c\right)dx$ (b) $\underset{a}{\overset{b}{\int }}f\left(x+c\right)dx$ (c) $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$ (d) $\underset{a-c}{\overset{b-c}{\int }}f\left(x\right)dx$

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Solution

## $\mathrm{Let}I=\underset{a+c}{\overset{b+c}{\int }}f\left(x\right)dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Let}x=t+c\phantom{\rule{0ex}{0ex}}⇒dx=dt\phantom{\rule{0ex}{0ex}}\mathrm{Also},\mathrm{if}x=a+c,t=a\phantom{\rule{0ex}{0ex}}\mathrm{if}x=b+c,t=b\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\phantom{\rule{0ex}{0ex}}I=\underset{a}{\overset{b}{\int }}f\left(t+c\right)dt\phantom{\rule{0ex}{0ex}}=\underset{a}{\overset{b}{\int }}f\left(x+c\right)dx\left(\because \underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f\left(t\right)dt\right)$ ​ Hence, the correct option is (b).

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