7.2 definite integral — Maths STD 12 Science — Question

$=\int \limits_0^1(x \times 0) d x+\int \limits_1^{\frac{3}{2}} x\left[x^2\right] d x$

$=\int \limits_1^{\frac{3}{2}} x\left[x^2\right] d x$

Let $x^2=t$ then $2 x d x=d t$. So,

$\int \limits_1^{\frac{3}{2}} x\left[x^2\right] d x=\frac{1}{2} \int \limits_1^{\frac{9}{4}}[t] d t$

$=\frac{1}{2} \int \limits_1^2[t] d t+\frac{1}{2} \int \limits_2^{\frac{9}{4}}[t] d t$

$=\frac{1}{2} \int \limits_1^2 1 d t+\frac{1}{2} \int \limits_2^{\frac{9}{4}} 2 d t$

$=\frac{1}{2}[t]_1^2+\frac{1}{2}[2 t]_2^{\frac{9}{4}}$

$=\frac{1}{2}(2-1)+\frac{1}{2}\left[2\left(\frac{9}{4}\right)-2(2)\right]$

$=\frac{1}{2}+\frac{1}{2}\left(\frac{9-8}{2}\right)$

$=\frac{3}{4}$