Question

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

A. $$\sqrt {{{\log }_e}\left( {3y - 2} \right)} $$
B. $$\root 3 \of {{{\log }_e}\left( {3y - 2} \right)} $$  
C. $$\root 3 \of {{{\log }_e}\left( {2 - 3y} \right)} $$
D. none of these
Answer :   $$\root 3 \of {{{\log }_e}\left( {3y - 2} \right)} $$
Solution :
$$\eqalign{ & y = \int {\frac{1}{3}{e^{{x^3}}}d\left( {{x^3}} \right)} = \frac{1}{3}{e^{{x^3}}} + c \cr & {\text{It passes through }}(0,{\text{ }}1) \cr & \therefore 1 = \frac{1}{3}{e^0} + c \cr & \therefore c = \frac{2}{3}.{\text{ Hence, }}y = \frac{1}{3}\left( {{e^{{x^3}}} + 2} \right) \cr & \therefore {e^{{x^3}}} = 3y - 2\,\,\,\,\,\,\,\,\,\,\therefore x = \root 3 \of {{{\log }_e}\left( {3y - 2} \right)} \cr} $$

Releted MCQ Question on
Calculus >> Indefinite Integration

Releted Question 1

The value of the integral $$\int {\frac{{{{\cos }^3}x + {{\cos }^5}x}}{{{{\sin }^2}x + {{\sin }^4}x}}dx} $$    is-

A. $$\sin \,x - 6\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
B. $$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} + c$$
C. $$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} - 6\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
D. $$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} + 5\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
Releted Question 2

If $$\int_{\sin \,x}^1 {{t^2}f\left( t \right)dt = 1 - \sin \,x} ,$$      then $$f\left( {\frac{1}{{\sqrt 3 }}} \right)$$   is-

A. $$\frac{1}{3}$$
B. $${\frac{1}{{\sqrt 3 }}}$$
C. $$3$$
D. $$\sqrt 3 $$
Releted Question 3

Solve this $$\int {\frac{{{x^2} - 1}}{{{x^3}\sqrt {2{x^4} - 2{x^2} + 1} }}dx} = ?$$

A. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{{x^2}}} + C$$
B. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{{x^3}}} + C$$
C. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{x} + C$$
D. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{2{x^2}}} + C$$
Releted Question 4

Let $$I = \int {\frac{{{e^x}}}{{{e^{4x}} + {e^{2x}} + 1}}dx,\,J = \int {\frac{{{e^{ - \,x}}}}{{{e^{ - \,4x}} + {e^{ - \,2x}} + 1}}dx.} } $$
Then for an arbitrary constant $$C,$$ the value of $$J-I$$  equals-

A. $$\frac{1}{2}\log \left( {\frac{{{e^{4x}} - {e^{2x}} + 1}}{{{e^{4x}} + {e^{2x}} + 1}}} \right) + C$$
B. $$\frac{1}{2}\log \left( {\frac{{{e^{2x}} + {e^x} + 1}}{{{e^{2x}} - {e^x} + 1}}} \right) + C$$
C. $$\frac{1}{2}\log \left( {\frac{{{e^{2x}} - {e^x} + 1}}{{{e^{2x}} + {e^x} + 1}}} \right) + C$$
D. $$\frac{1}{2}\log \left( {\frac{{{e^{4x}} + {e^{2x}} + 1}}{{{e^{4x}} - {e^{2x}} + 1}}} \right) + C$$

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Indefinite Integration


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