Question

Evaluate the definite integral: \(∫^1_{-1}(x+1)dx\)

Answer 1

The correct answer is: 2
Let \(I=∫^1_{-1}(x+1)dx\)
\(∫(x+1)dx=\frac{x^2}{2}+x=F(x)\)
By second fundamental theorem of calculus,we obtain
\(I=F(1)-F(-1)\)
\(=(\frac{1}{2}+1)-(\frac{1}{2}-1)\)
\(=\frac{1}{2}+1-\frac{1}{2}+1\)
\(=2\)