$\underset{a+c}{\overset{b+c}{\int }}f\left(x\right)dx$ 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) $\underset{a-c}{\overset{b-c}{\int }}f\left(x\right)dx$

$$\begin{gathered} \mathrm{Let}I=\underset{a+c}{\overset{b+c}{\int }}f\left(x\right)dx \\ \mathrm{Let}x=t+c \\ \Rightarrow dx=dt \\ \mathrm{Also},\mathrm{if}x=a+c,t=a \\ \mathrm{if}x=b+c,t=b \\ \mathrm{Thus}, \\ I=\underset{a}{\overset{b}{\int }}f\left(t+c\right)dt \\ =\underset{a}{\overset{b}{\int }}f\left(x+c\right)dx\left(\because \underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f\left(t\right)dt\right) \end{gathered}$$

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Hence, the correct option is (b).