Question
Evaluate the definite integral $\int_{-1}^{1}(x+1) d x$

Answer

Let $I=\int_{-1}^{1}(x+1) d x$ 
$I=\int_{-1}^{1}(x+1) d x$ 
$\Rightarrow \mathrm{I}=\int_{-1}^{1} \mathrm{x} \mathrm{dx}+\int_{-1}^{1} 1 \cdot \mathrm{dx} ~~~~~\left[As \int \mathrm{x}^{\mathrm{n}} \mathrm{d} \mathrm{x}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1}\right]$ 
$\Rightarrow I=\left[\frac{x^{2}}{2}\right]_{-1}^{1}+[x]_{-1}^{1}$ 
$\Rightarrow \mathrm{I}=\left[\frac{1^{2}}{2}-\frac{(-1)^{2}}{2}\right]+[1-(-1)]$ 
$\Rightarrow I=\left[\frac{1}{2}-\frac{1}{2}\right]+[1+1]=0+2$ 
$\Rightarrow$ I = 2
$\int_{-1}^{1}(x+1) d x=2$

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