Download App

Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int_\limits{\frac{1}{3}}^{1}\frac{\big(\text{x}-\text{x}^3\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
✓

Answer

Let $\text{I}=\int_\limits{\frac{1}{3}}^{1}\frac{\big(\text{x}-\text{x}^3\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$$=\int_\limits{\frac{1}{3}}^{1}\frac{\Bigg[\text{x}^3\Big(\frac{\text{x}}{\text{x}^3}-1\Big)\Bigg]^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\text{x}\big(\frac{1}{\text{x}^2}-1\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\text{x}\big(\frac{1}{\text{x}^2}-1\big)^{\frac{1}{3}}}{\text{x}^3}\text{ dx}$
Put $\Big(\frac{1}{\text{x}^2}-1\Big)=\text{Z}$$\therefore\ -\frac{2}{\text{x}^3}\text{ dx}=\text{dz}$
$\Rightarrow\frac{\text{dx}}{\text{x}^3}=-\frac{\text{dz}}{2}$
When $\text{x}\rightarrow\frac{1}{3},\text{z}\rightarrow8$ When $\text{x}\rightarrow1,\text{z}\rightarrow0$$\therefore\ \text{I}=-\frac{1}{2}\int^\limits0_8\text{z}^{\frac{1}{3}}\text{ dz}$
$=-\frac{1}{2}\times\Bigg[\frac{\text{z}^{\frac{4}{3}}}{\frac{4}{3}}\Bigg]^0_8$
$=-\frac{3}{8}\Big[0-(8)^{\frac{4}{3}}\Big]$
$=-\frac{3}{8}\times(-16)$
$=6$

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

Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}\text{ on }[1,3]$
→
If $\text{x}-\text{e}^{\frac{\text{x}}{\text{y}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}$
→
Find the intervals in which the following functions are increasing or decreasing.
$f(x) = (x - 1)(x - 2)^2$
→
Solve the following differential equation:
$(\text{x + y})(\text{dx}-\text{dy})=\text{dx + dy}$
→
If $\text{A}=\begin{bmatrix}0&0\\4&0\end{bmatrix},$ find $A^{16}.$
→
Draw the rough sketch of $\frac{\text{x}^{2}}{4}+\frac{\text{y}^{2}}{9}=1$ and evaluate the area of the region under the area the curve and the line x-axis.
→
Evaluate the following integrals:$\int\limits^2_{-2}\frac{3\text{x}^3+2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
→
Evaluate the following integrals:
$\int\text{x}^3\tan^{-1}\text{x dx}$
→
Show that $\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}=0$ where a, b, c are in A.P.
→
Determine the equation of the line passing through the points (1, 2, -4) and perpendicular to the lines $\frac{\text{x}-8}{8}=\frac{\text{y}+9}{-16}=\frac{\text{z}-10}{7}$ and $\frac{\text{x}-15}{3}=\frac{\text{y}-29}{8}=\frac{\text{z}-5}{-5}.$
→