Download App

INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Evaluate the definite integral in Exercise:
$\int^{4}_{1}[\text{x}-1|+|\text{x}-2|+|\text{x}-3|]\text{dx}$
✓

Answer

$\text{Let I}=\int^{4}_{1}[\text{x}-1|+|\text{x}-2|+|\text{x}-3|]\text{dx}$
$\Rightarrow\text{I}=\int^{4}_{1}\text{x}-1|\text{dx}+\int^{4}\limits_{1}|\text{x}-2|\text{dx}+\int^{4}\limits_{1}|\text{x}-3|\text{dx}$
$\text{I}=\text{I}_{1}+\text{I}_{2}+\text{I}_{3}$
where,$\text{I}_{1}=\int^{4}\limits_{1}\text{x}-1|\text{dx},\text{I}_{1}=\int^{4}\limits_{1}|\text{x}-2|\text{dx},$and $\text{I}_{3}=\int^{4}\limits_{1}\text{x}-3|\text{dx}$
$\text{I}_{1}=\int^{4}_{1}|\text{x}-1|\text{dx}$
$(\text{x}-1)\geq0$ for $1\le\text{x}\le4$
$\therefore\text{I}_{1}=\int^{4}\limits_{1}(\text{x}-1)\text{dx}$
$\Rightarrow\text{I}_{1}=\bigg[\frac{\text{x}^{2}}{\text{2}}-\text{x}\bigg]^{4}_{1}$
$\Rightarrow\text{I}_{1}=\bigg[8-4-\frac{1}{2}+1\bigg]=8-4+\frac{1}{2}$
$\text{I}_{2}=\int^{4}\limits_{1}|\text{x}-2|\text{dx}$
$\text{x}-2\ge0$ for $2\le\text{x}\le4$ and $\text{x}-2\le0$ for $1\le\text{x}\le2$
$\therefore\ \text{I}_{2}=\int^{2}\limits_{1}(2-\text{x)}\text{dx}+\int^{4}\limits_{2}(\text{x}-2)\text{dx}$
$\Rightarrow\ \text{I}_{2}=\bigg[2\text{x}-\frac{\text{x}^{2}}{2}\bigg]^{2}_{1}+\bigg[\frac{\text{x}^{2}}{2}-2\text{x}\bigg]^{4}_{2}$
$\Rightarrow\ \text{I}_{2}=\bigg[4-2-2+\frac{1}{2}\bigg]+\bigg[8-8-2+4\bigg]$
$\Rightarrow\ \text{I}_{2}=\frac{1}{2}+2=\frac{5}{2}$
$\text{I}_{3}=\int^{4}\limits_{1}|\text{x}-3|\text{dx}$
$\text{x}-3\ge0$ for $3\le\text{x}\le4$ and $\text{x}-3\le$ for $1\le\text{x}\le3$
$\therefore\text{I}_{3}=\int^{3}\limits_{1}(3-\text{x})\text{dx}+\int^{4}\limits_{3}(\text{x}-3)\text{dx}$
$\Rightarrow\ \text{I}_{3}=\bigg[3\text{x}-\frac{\text{x}^{2}}{2}\bigg]^{3}_{1}+\bigg[\frac{\text{x}^{2}}{2}-3\text{x}\bigg]^{4}_{3}$
$\Rightarrow\ \text{I}_{3}=\bigg[9-\frac{9}{2}-3+\frac{1}{2}\bigg]+\bigg[8-12-\frac{9}{2}+9\bigg]$
$\Rightarrow\ \text{I}_{3}=\big[6-4\big]+\bigg[\frac{1}{2}\bigg]=\frac{5}{2}$
From equation (1),(2),(3), and (4), we obtain
$\text{I}=\frac{9}{2}+\frac{5}{2}+\frac{5}{2}=\frac{19}{2}$

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:
$\int\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$
→
Evaluate the following intregals:
$\int\frac{1}{\sin\text{x}-\cos\text{x}}\ \text{dx}$
→
$\begin{vmatrix}\text{b}+\text{c}&\text{a}&\text{a}\\\text{b}&\text{c}+\text{a}&\text{b}\\\text{c}&\text{c}&\text{a}+\text{b}\end{vmatrix}=4\text{abc}$
→
If $\sec\Big(\frac{\text{x}+\text{y}}{\text{x}-\text{y}}\Big)=\text{a}$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
→
Maximum Z = 4x + 3y
Subject to
$3\text{x}+4\text{y}\leq24$
$8\text{x}+6\text{y}\leq48$
$\text{x}\leq5$
$\text{y}\leq6$
$\text{x},\text{y}\geq0$
→
Solve the following equation for x:
$\tan^{-1}\frac{1}{4}+2\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{6}+\tan^{-1}\frac{1}{\text{x}}=\frac{\pi}{4}$
→
Evaluate the following definite integrals:
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}$
→
Prove that the line of section of the planes $5x + 2y - 4z + 2 = 0$ and $2x + 8y + 2z - 1 = 0$ is parallel to the plane $4x - 2y - 5z - 2 = 0.$
→
Evaluate the following integrals:
$\int\sqrt{2\text{x}-\text{x}^2}\text{dx}$
→
If $\text{A}=\begin{bmatrix}2&3\\-1&0\end{bmatrix},$ show that $A^2 - 2A + 3I_2 = 0$.
→