Download App

INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Evaluate the following integrals:
$\int(2\text{x}+5)\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}$
✓

Answer

$\text{I}=\int(2\text{x}+5)\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}$Let $(2\text{x}+5)=\text{A}\frac{\text{d}}{\text{dx}}(10-4\text{x}-3\text{x}^2)+\text{B}$
$\Rightarrow(2\text{x}+5)=\text{A}(-4-6\text{x})+\text{B}$
$\Rightarrow(2\text{x}+5)=-6\text{A}\text{x}+(\text{B}-4\text{A})$
$\Rightarrow2=-6\text{A}\text{ and }(\text{B}-4\text{A})=5$
$\Rightarrow\text{A}=-\frac{1}{3}\text{ and }\text{B}=\frac{11}{3}$
$\Rightarrow(2\text{x}+5)=-\frac{1}{3}(-4-6\text{x})+\frac{11}{3}$
$\Rightarrow\text{I}=-\frac{1}{3}(-4-6\text{x})\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}\\+\frac{11}{3}\int\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}$
Let $\text{I}=-\frac{1}{3}\text{I}_1+\frac{11}{3}\text{I}_2\ \dots(1)$
Now,
$\text{I}_1=\int(-4-6\text{x})\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}$
Let $(10-4\text{x}-3\text{x}^2)=\text{t},\text{ or, }(-4-6\text{x})\text{dx = dt}$
$\Rightarrow\text{I}_1=\int\sqrt{\text{t}}\text{dt}$
$=\frac{2}{3}\text{t}^{\frac{3}{2}}+\text{c}_1$
$\Rightarrow\text{I}_1=\frac{2}{3}(10-4\text{x}-3\text{x}^2)^{\frac{3}{2}}+\text{c}_1$
$\text{I}_2=\int\sqrt{(10-4\text{x}-3\text{x}^2)}\text{dx}$
$=\int\sqrt{3\Big(\frac{10}{3}-\frac{4}{3}\text{x}-\text{x}^2\Big)}\text{dx}$
$=\sqrt3\int\sqrt{\Big(\frac{26}{9}-\frac{4}{9}-\frac{4}{3}\text{x}-\text{x}^2\Big)}\text{dx}$
$=\sqrt3\int\sqrt{\Big[\Big(\frac{\sqrt{26}}{3}\Big)^2-\Big(\frac{4}{9}+\frac{4}{3}\text{x}-\text{x}^2\Big)\Big]}\text{dx}$
$=\sqrt3\int\sqrt{\Big[\Big(\frac{\sqrt{26}}{3}\Big)^2-\Big(\text{x}+\frac{2}{3}\Big)^2\Big]}\text{dx}$
$=\sqrt3\sin\Bigg(\frac{\text{x}+\frac{2}{3}}{\frac{\sqrt{26}}{3}}\Bigg)+\text{c}_2$
$=\sqrt3\sin\Big(\frac{3\text{x}+2}{\sqrt{26}}\Big)+\text{c}_2$
Using (1), we get
$\text{I}=-\frac{1}{3}\times\frac{2}{3}(10-4\text{x}-3\text{x}^2)^{\frac{3}{2}}\\+\frac{11}{3}\times\sqrt3\sin\Big(\frac{3\text{x}+2}{\sqrt{26}}\Big)+\text{C}$
$\therefore\ \text{I}=-\frac{2}{9}(10-4\text{x}-3\text{x}^2)^{\frac{3}{2}}+\frac{11\sqrt3}{3}\sin\Big(\frac{3\text{x}+2}{\sqrt{26}}\Big)+\text{C}$

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

Prove that:
$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z}).$
→
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}\text{ on }[0,\pi]$
→
A bag has 4 red and 6 black balls and other bag has 3 red and 5 black balls. One of the two bags, is selected at random and a ball is drawn at random. The ball drawn is found to be red. What is the probability that ball comes out from second bag?
→
Prove that:
$\cos^{-1}\frac{12}{13}=\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{56}{65}$
→
Differentiate the following functions with respect to x:
$\tan^{-1}\bigg[\frac{\text{x}^\frac{1}{3}+\text{a}^{\frac{1}{3}}}{1-(\text{ax})^\frac{1}{3}}\bigg]$
→
Evaluate the following intregals:
$\int\frac{\text{x}^3}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}$
→
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes.
→
Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2 R}{\sqrt{3}}$ Also find the maximum volume.
→
Evaluate the following integrals:
$\int\limits_{0}^{1}\text{x}\log(1+2\text{x})\text{dx}$
→
Draw a rough sketch of the region $\{(x, y) : y^2 < 5x, 5x^2 + < 36\}$ and find the area by the region using mwthod of integration.
→