Mathematics

# Let the equation of a curve passing through the point $\displaystyle \left ( 0, 1 \right )$ be given by $\displaystyle y=\int x^{2}.e^{x^{3}}dx.$ If the equation of the curve is written in the form $\displaystyle x=f\left ( y \right )$ then $\displaystyle f\left ( y \right )$ is

$\displaystyle \sqrt[3]{\log _{e}\left ( 3y-2 \right )}$

##### SOLUTION

Substitute $x^3=t$

$\Rightarrow 3x^2dx=dt$

The equation becomes $\displaystyle y=\frac { 1 }{ 3 } \int { { e }^{ t }dt } \\ \displaystyle =\frac { 1 }{ 3 } { e }^{ t }+c\\ \displaystyle =\frac { 1 }{ 3 } { e }^{ { x }^{ 3 } }+c$

Given that the curve passes through $(0,1)$. Using this we get

$c=\dfrac{2}{3}$

So, equation becomes $\displaystyle 3y=e^{x^3}+2$

$\Rightarrow e^{x^3}=3y-2$

$\Rightarrow x^3 = \ln{(3y-2)}$

$\displaystyle \Rightarrow x=\sqrt[3]{ \ln{(3y-2)}}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Hard
let $f(x)=(x+1)^{2}+\frac{1}{x}$ then the value of $\int_{-2}^{1}f(x)(-x)dx$
• A. is equal to $-\frac{81}{10}$
• B. is equal to $\frac{81}{10}$
• C. does not exist
• D. is equal to $\dfrac{-15}{4}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
The value of $\int\dfrac{{x^{2}}}{{(x^{2}+a^{2})(x^{2}+b^{2})}}dx$ is
• A. $\dfrac{1}{b^{2}-a^{2}}[b\tan^{-1}]-a\tan^{-1}\dfrac{x}{a}]+C$
• B. $\dfrac{1}{b^{2}-a^{2}}[a\tan^{-1}]-b\tan^{-1}\dfrac{x}{a}]+C$
• C. $\dfrac{1}{b^{2}-a^{2}}[b\tan^{-1}]+a\tan^{-1}\dfrac{x}{a}]+C$
• D. $none\ of\ these$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate the following integral
$\int { \cfrac { { e }^{ 2x } }{ { e }^{ 2x }-2 } } dx\quad$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
$I = \int \dfrac {1}{x(x^{6}+1)}dx$

If $f,g,h$ be continuous functions on $[0,a]$ such that $f(a-x)=-f(x),g(a-x)=g(x)$ and $3h(x)-4h(a-x)=5$ then  $\displaystyle \int_0^a f(x)g(x)h(x)dx=0$