Download App

Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\limits^2_{-2}\frac{3\text{x}^3+2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
✓

Answer

Let $\text{I}=\int\limits^2_{-2}\frac{3\text{x}^3+2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$$=\int\limits^2_{-2}\frac{3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}+\int\limits^2_{-2}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
$=\text{I}_1+\text{I}_2$
Consider $\text{f(x)}=\int\limits^2_{-2}\frac{3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}$$\text{f}(-\text{x})=\int\limits^2_{-2}\frac{3(-\text{x}^3)}{(-\text{x}^2)+|-\text{x}|+1}\text{ dx}=\int\limits^2_{-2}\frac{-3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}=-\int\limits^2_{-2}\frac{3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}=-\text{f(x)}$
$\therefore\ \text{I}_1=\int\limits^2_{-2}\frac{3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}=0$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
Now, Consider$\text{g(x)}=\int\limits^{2}_{-2}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
$\text{g}(-\text{x})=\int\limits^{2}_{-2}\frac{2|-\text{x}|+1}{(-\text{x}^2)+|-\text{x}|+1}\text{ dx}=\text{g(x)}=\int\limits^{2}_{-2}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}=\text{g}(\text{x)}$
$\therefore\ \text{I}_2=\int\limits^{2}_{-2}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
$=2\int\limits^{2}_{0}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
$=2\int\limits^{2}_{0}\frac{2\text{x}+1}{\text{x}^2+\text{x}+1}\text{ dx}$ $\begin{bmatrix}|\text{x}|\begin{cases}\text{x},&\text{x}\geq0\\-\text{x},&\text{x}<0\end{cases}\end{bmatrix}$
$=2\times\Big[\log(\text{x}^2+\text{x}+1)\Big]^2_0$
$=2\times\big(\log7-\log1\big)$
$=2\times\big(7-0\big)$
$=2\log7$
$\therefore\ \text{I}=\text{I}_1+\text{I}_2$
$=0+2\log7$
$=2\log7$

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 "Solve the Following Question.(5 Marks)" from Definite Integrals→Definite Integrals — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix}7 & 1 \\4 & -3 \end{bmatrix}$
→
The function $y=a \log x+b x^2+x$ has extreme values at $x=1$ and $x=2$. Find $a$ and $b$.
→
Form the differential equation of the family of curves represented by the equation (a being the perimeter):$(\text{x}-\text{a})^2+2\text{y}^2=\text{a}^2$
→
If $x = a \cos \theta, y = b \sin \theta$, show that $a^2\left[y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2\right]+b^2=0$
→
If $\text{y}=\cos^{-1}\Big\{\frac{2\text{x}-3\sqrt{1-\text{x}^2}}{\sqrt{13}}\Big\},$ find $\frac{\text{dy}}{\text{dx}}.$
→
Find the vector and Cartesian equations of the plane that passes through the point (5, 2, -4) and is perpendicular to the line with direction ratios 2, 3, -1.
→
Find the percentage error in calculating the surface area of a cubical box if an error of 1% is made in measuring the lengths of edges of the cube.
→
Solve the following initial value problems:
$(\text{x}^2+\text{y}^2)\text{dx}=2\text{xy dy, y}(1)=0$
→
Maximize z = 60x + 50y subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0
→
Find the area of the bounded by the curve $\text{y}=\frac{\pi}{2}+2\sin^{2}\text{x}$ x-axis and the area between x-axis, the curve and the ordinates $\text{x}=0, \text{x}=\pi$.
→