Question Bank [2022] — Maths STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]3 Marks
Question
Evaluate: $\int_0^1 t ^2 \sqrt{1- t } dt$
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Answer
$\text { Let } I =\int_0^1 t ^2 \sqrt{1- t } dt$
$=\int_0^1(1- t )^2 \sqrt{1-(1- t )} dt \quad \ldots . . .\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$=\int_0^1\left(1-2 t+t^2\right) \sqrt{t} d t$
$=\int_0^1\left(t^{\frac{1}{2}}-2 t^{\frac{3}{2}}+t^{\frac{5}{2}}\right) d t$
$=\int_0^1 t^{\frac{1}{2}} d t-2 \int_0^1 t^{\frac{3}{2}} d t+\int_0^1 t^{\frac{5}{2}} d t$
$=\left[\frac{ t ^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^1-2\left[\frac{ t ^{\frac{5}{2}}}{\frac{5}{2}}\right]_0^1+\left[\frac{ t ^{\frac{7}{2}}}{\frac{7}{2}}\right]_0^1$
$=\frac{2}{3}\left(1^{\frac{3}{2}}-0\right)-\frac{4}{5}\left(1^{\frac{5}{2}}-0\right)+\frac{2}{7}\left(1^{\frac{7}{2}}-0\right)$
$=\frac{2}{3}-\frac{4}{5}+\frac{2}{7}$
$=\frac{70-84+30}{105}$
$\therefore I =\frac{16}{105}$
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