Download App

INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Evaluate the following definite integral as limit of sum:$\int_\limits{0}^{4}\text{(x}+\text{e}^{2\text{x}})\ \text{dx}$
✓

Answer

$\text{we}\ \text{know}\ \text{that}\ \int\limits_{\text{a}}^{\text{b}}\text{f}\text{(x)}\ \text{dx}=\lim\limits_{\text{h}\rightarrow0}\ \text{h}\big[\text{f}\ \text{(a)}+\text{f}\text{(a+h)}+\text{f}\text{(a}+2\text{h})+.....+\text{f}\text{(a}+\text{(n}-1)\text{h)}\big] $ $\text{where}\ \text{nh}=\text{b}-\text{a}$ $\text{Here},\ \text{a}=0,\text{b}=4,\text{nh}=4\ \text{and}\ \text{f}\ \text{(x)}=\text{x}+\text{e}^{2\text{x}}$ $\therefore\ \ \int\limits_{0}^{4}\text{(x+e}^{2\text{h}})\ \text{dx}=\lim\limits_{\text{h}\rightarrow0}\ \text{h}\bigg[1+\text{(h}+\text{e}^{2\text{h}})+\text{(2h}+\text{e}^{4\text{h}})+....+\big(\text{(n}-1)\text{h+e}^{2\text{(n-1)}\text{h}}\big)\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{(h}+2\text{h}+.....+\text{(n}-1)\text{h)}+\big(1+\text{e}^{2\text{h}}+\text{e}^{4\text{h}}+....+\text{e}^{2(\text{n}-1)\text{h}}\big)\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{h}(1+2+......+\text{(n}-1)+\text{a}\bigg(\frac{\text{r}^{\text{n}}-1}{\text{r}-1}\bigg)\bigg]=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{h.h}\frac{\text{n}\text{(n}-1)}{2}+\frac{1\big(\text{e}^{2\text{h}^{\text{n}}}\big)-1}{\text{e}^{2\text{h}}-1}\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}=\bigg[\frac{\text{nh}\text{(nh}-\text{h)}}{2}+\frac{\text{h}\big(\big(\text{e}^{2\text{nh}}\big)-1\big)}{\text{e}^{2\text{h}}-1}\bigg] =\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{4(4-\text{h)}}{2}+\frac{\text{h}\big(\big(\text{e}^{2.4}\big)-1\big)}{\text{e}^{2\text{h}}-1}\bigg]$$=\frac{4(4-0)}{2}+\big(\text{e}^{8}-1\big)\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{e}^{2\text{h}}-1}=8\big(\text{e}^{8}-1\big)\frac{1}{2}\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}}{\text{e}^{2\text{h}}-1}=8+\frac{\big(\text{e}^{8}-1\big)}{2}$
$=\bigg[\because\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\text{e}^{\text{x}}-1}=1\bigg]$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from INTEGRALS→INTEGRALS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Evaluate the following integrals:
$\int\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
→
Solve the following systems of linear equations by cramer's rule:
2x - 3y - 4z = 29,
-2x + 5y - z = -15,
3x - y + 5z = -11
→
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\text{y}=\sin^{-1}\Bigg(\frac{2\text{x}}{1+\text{x}^{2}}\Bigg)$
→
If $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$ and $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then prove that $A^2 - A + 2I = O$.
→
Show that $\text{f}(\text{x})=\tan^{-1}(\sin\text{x}+\cos\text{x})$ is a decreasing function on the interval $\Big(\frac{\pi}{4},\frac{\pi}{2}\Big).$
→
If $\text{y}=\log\sqrt{\text{x}+1}+\sqrt{\text{x}-1},$ show that $\sqrt{\text{x}^2-1}\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\text{y}.$
→
Find the particular solution of the differential equation $\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big)\frac{\text{dy}}{\text{dx}}=\text{y}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{x},$ given that when $\text{x}=1,\text{y}=\frac{\pi}4$.
→
Find the area of the region included between $y^2 = 9x$ and $y = x$.
→
Using differentials, find the approximate value of $\sqrt{49.5}$
→
Determine the intervals in which the function $f (x) = x^4 - 8x^3 + 22x^2 - 24x + 21$is strictly increasing or strictly decreasing.
→