Question

Evaluate the following integrals as limit of sums:

${\int }_1^3\left(3x^2+1\right)dx$ [CBSE 2014]

Answer 1

We have,

${\int }_a^{b}f\left(x\right)dx=\underset{h\to 0}{\lim }\left\{f\left(a\right)+f\left(a+h\right)+f\left(a+2h\right)+...+f\left[\left(a+\left(n-1\right)h\right)\right]\right\}$

Here, a = 1, b = 3, f(x) = 3x² + 1 and $h=\frac{3-1}{n}=\frac{2}{n}\Rightarrow nh=2$

$$\begin{gathered} \therefore {\int }_1^3\left(3x^2+1\right)dx \\ =\underset{h\to 0}{\lim }h\left\{f\left(1\right)+f\left(1+h\right)+f\left(1+2h\right)+...+f\left[1+\left(n-1\right)h\right]\right\} \\ =\underset{h\to 0}{\lim }h\left\{\left[3\times 1^2+1\right]+\left[3\times {\left(1+h\right)}^2+1\right]+\left[3\times {\left(1+2h\right)}^2+1\right]+...+\left[3\times {\left(1+\left(n-1\right)h\right)}^2+1\right]\right\} \\ =\underset{h\to 0}{\lim }h\left\{3\left[1+\left(1+2h+h^2\right)+\left(1+4h+2^2{h}^2\right)+...+\left(1+2\left(n-1\right)h+{\left(n-1\right)}^2{h}^2\right)\right]+n\right\} \\ =\underset{h\to 0}{\lim }h\left\{3\left[n+2\left(1+2+...+\left(n-1\right)\right)h+\left(1^2+2^2+...+{\left(n-1\right)}^2\right)h^2\right]+n\right\} \\ =\underset{h\to 0}{\lim }h\left[4n+6\times \frac{n\left(n-1\right)}{2}h+3\times \frac{\left(n-1\right)n\left(2n-1\right)}{6}{h}^2\right] \end{gathered}$$
$$\begin{gathered} =\underset{h\to 0}{\lim }\left[4nh+6\times \frac{nh\left(nh-h\right)}{2}+3\times \frac{\left(nh-h\right)nh\left(2nh-h\right)}{6}\right] \\ =\underset{h\to 0}{\lim }\left[4nh+3\times nh\left(nh-h\right)+3\times \frac{\left(nh-h\right)nh\left(2nh-h\right)}{6}\right] \\ =\underset{h\to 0}{\lim }\left[4\times 2+3\times 2\times \left(2-h\right)+3\times \frac{\left(2-h\right)\times 2\times \left(2\times 2-h\right)}{6}\right] \\ =8+6\times \left(2-0\right)+\frac{\left(2-0\right)\times 2\times \left(4-0\right)}{2} \\ =8+12+8 \\ =28 \end{gathered}$$