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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration5 Marks
Question
Evaluate : $\int_0^4\left(x-x^2\right) \cdot d x$
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Answer

$
\begin{aligned}
& \int_0^4\left(x-x^2\right) \cdot d x=\int_a^b f(x) d x \\
& f(x)=x-x^2 \quad a=0 ; b=4 \\
& \Rightarrow \quad f(a+r h)=f(0+r h) \quad \text { and } \\
& =f(r h) \\
& =(r h)-(r h)^2 \\
& h=\frac{b-a}{n} \\
& =r h-r^2 h^2 \\
& h=\frac{4-0}{n} \\
& \therefore \quad n h=4 \\
&
\end{aligned}
$
We know $\int_a^b f(x) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot[f(a+r h)]$
$
\begin{aligned}
\therefore \quad \int_0^4\left(x-x^2\right) \cdot d x & =\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot\left(r h-r^2 h^2\right) \\
& =\lim _{n \rightarrow \infty} \sum_{r=1}^n\left(r h^2-r^2 h^3\right) \\
& =\lim _{n \rightarrow \infty}\left(h^2 \cdot \sum_{r=1}^n r-h^3 \cdot \sum_{r=1}^n r^2\right) \\
& =\lim _{n \rightarrow \infty}\left[h^2\left(\frac{n(n+1)}{2}\right)-h^3\left(\frac{n(n+1)(2 n+1)}{6}\right)\right] \\
& =\lim _{n \rightarrow \infty}\left[\frac{h^2 \cdot n \cdot n\left(1+\frac{1}{n}\right)}{2}-\frac{h^3 \cdot n \cdot n\left(1+\frac{1}{n}\right) n\left(2+\frac{1}{n}\right)}{6}\right] \\
& =\lim _{n \rightarrow \infty}\left[\frac{(n h)^2\left(1+\frac{1}{n}\right)}{2}-\frac{(n h)^3\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}\right] \\
& =\lim _{n \rightarrow \infty}\left[\frac{(4)^2\left(1+\frac{1}{n}\right)}{2}-\frac{(4)^3\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}\right] \\
& =\frac{(4)^2 \cdot(1+0)}{2}-\frac{(4)^3(1+0)(2+0)}{6} \\
& =-\frac{(64)(2)}{6}
\end{aligned}
$

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