JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 22)
If $$\int \, $$x5.e$$-$$4x3 dx = $${1 \over {48}}$$e$$-$$4x3 f(x) + C, where C is a constant of inegration, then f(x) is equal to -
$$-$$2x3 $$-$$ 1
$$-$$ 2x3 + 1
4x3 + 1
$$-$$4x3 $$-$$ 1
Explanation
$$\int {{x^5}} .{e^{ - 4{x^3}}}\,dx = {1 \over {48}}{e^{ - 4{x^3}}}f\left( x \right) + c$$
Put $${x^3} = t$$
$$3{x^2}\,dx = dt$$
$$\int {{x^3}.{e^{ - 4{x^3}}}.\,{x^2}} dx$$
$${1 \over 3}\int {t.{e^{ - 4t}}dt} $$
$${1 \over 3}\left[ {t.{{{e^{ - 4t}}} \over { - 4}} - \int {{{{e^{ - 4t}}} \over { - 4}}dt} } \right]$$
$$ - {{{e^{ - 4t}}} \over {48}}\left[ {4t + 1} \right] + c$$
$${{ - {e^{ - 4{x^3}}}} \over {48}}\left[ {4{x^3} + 1} \right] + c$$
$$ \therefore $$ $$f(x) = - 1 - 4{x^3}$$
Put $${x^3} = t$$
$$3{x^2}\,dx = dt$$
$$\int {{x^3}.{e^{ - 4{x^3}}}.\,{x^2}} dx$$
$${1 \over 3}\int {t.{e^{ - 4t}}dt} $$
$${1 \over 3}\left[ {t.{{{e^{ - 4t}}} \over { - 4}} - \int {{{{e^{ - 4t}}} \over { - 4}}dt} } \right]$$
$$ - {{{e^{ - 4t}}} \over {48}}\left[ {4t + 1} \right] + c$$
$${{ - {e^{ - 4{x^3}}}} \over {48}}\left[ {4{x^3} + 1} \right] + c$$
$$ \therefore $$ $$f(x) = - 1 - 4{x^3}$$
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