Question

Evaluate the definite integrals: $\int_{-1} ^{1} {(x+1)} \, dx$

Answer 1

Answer:

2

Step-by-step explanation:

Formula used:

$\int{x^n}\:dx=\frac{x^{n+1}}{n+1}+c$

Now,

$\int\limits^1_{-1}[x+1]dx\\\\=[\frac{x^2}{2}+x]^1_{-1}\\\\=[\frac{1}{2}+1]-[\frac{1}{2}-1]\\\\=\frac{1}{2}+1-\frac{1}{2}+1\\\\=2$