Question

Evaluate the definite integral: $∫_1^0\frac{dx}{1+x^2}$

Answer 1

Let I=$∫_1^0\frac{dx}{1+x^2}$

$∫=\frac{dx}{1+x^2}=tan^-1x=F(x)$

By second fundamental theorem of calculus,we obtain

$I=F(1)-F(0)$

$=tan-1(1)-tan-1(0)$

$=\frac{\pi}{4}$