Question 13.7: Evaluate (a) ∫²1 x² + 1 dx  (b) ∫¹2 x² + 1 dx (c) ∫^π 0 sin ......

(a) Let I stand for  \int_1^2 x^2+1 d x.

I=\int_1^2 x^2+1 \mathrm{~d} x=\left[\frac{x^3}{3}+x\right]_1^2

The integral is now evaluated at the upper and lower limits. The difference gives the value required.

I=\left(\frac{2^3}{3}+2\right)-\left(\frac{1^3}{3}+1\right)=\frac{8}{3}+2-\frac{4}{3}=\frac{10}{3}

(b) Because interchanging the limits of integration changes the sign of the integral, we find

\int_2^1 x^2+1 \mathrm{~d} x=-\int_1^2 x^2+1 \mathrm{~d} x=-\frac{10}{3}

(c) \int_0^\pi \sin x d x=[-\cos x]_0^\pi=(-\cos \pi)-(-\cos 0)=1-(-1)=2

Figure 13.10 illustrates this area.