Mathematics - Definite Integration Question with Solution | TestHub

Let $I={\int }_{-1}^1[x[1+\sin \pi x]+1]dx$.

Now, if $-1<x<0$, then $[1+\sin \pi x]=0$.

And if $0<x<1$, then $[1+\sin \pi x]=1$.

Therefore:

$$I={\int }_{-1}^{0}1\cdot dx+{\int }_0^1[x+1]dx$$

$$I={\int }_{-1}^{0}1\cdot dx+{\int }_0^1[x]dx+{\int }_0^{1}1\cdot dx$$

$$I=[0-(-1)]+0+(1-0)=2$$