MCQ
$\mathop \smallint \limits_0^{1.5} x\left[ {{x^2}} \right]dx = $
- A$0$
- B$\frac{3}{2}$
- ✓$\frac{3}{4}$
- D$\frac{5}{4}$
✓
Answer
Correct option: C.
$\frac{3}{4}$
c
$\int \limits_0^{1.5} x\left[x^2\right] d x=\int \limits_0^1 x\left[x^2\right] d x+\int \limits_1^{\sqrt{2}} x\left[x^2\right] d x+\int \limits_{\sqrt{2}}^{1.5} x\left[x^2\right] d x$
$\int \limits_0^{1.5} x\left[x^2\right] d x=\int \limits_0^1 x\left[x^2\right] d x+\int \limits_1^{\sqrt{2}} x\left[x^2\right] d x+\int \limits_{\sqrt{2}}^{1.5} x\left[x^2\right] d x$
$=\int \limits_0^1(x \times 0) d x+\int \limits_1^{\frac{3}{2}} x\left[x^2\right] d x$
$=\int \limits_1^{\frac{3}{2}} x\left[x^2\right] d x$
Let $x^2=t$ then $2 x d x=d t$. So,
$\int \limits_1^{\frac{3}{2}} x\left[x^2\right] d x=\frac{1}{2} \int \limits_1^{\frac{9}{4}}[t] d t$
$=\frac{1}{2} \int \limits_1^2[t] d t+\frac{1}{2} \int \limits_2^{\frac{9}{4}}[t] d t$
$=\frac{1}{2} \int \limits_1^2 1 d t+\frac{1}{2} \int \limits_2^{\frac{9}{4}} 2 d t$
$=\frac{1}{2}[t]_1^2+\frac{1}{2}[2 t]_2^{\frac{9}{4}}$
$=\frac{1}{2}(2-1)+\frac{1}{2}\left[2\left(\frac{9}{4}\right)-2(2)\right]$
$=\frac{1}{2}+\frac{1}{2}\left(\frac{9-8}{2}\right)$
$=\frac{3}{4}$
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