- f x dx - x - then - x 5 f x 3 dx is equal toA- 1-3- x 3- x 3- x 2- x 3 dx - cB- 1-3 x 3-x3- x 3- x 3 dx - c- 1-3-3-x3- x 2- x 3 dx - cD- 1-3- x 3- x 3- x 3- x 3 dx - c

Given, ∫ f(x)dx=ψ(x) I=∫ x^5f(x^3)dx put x^3=t⇒ x^2dx=13 dt ⇒ I=13∫ tf(t)dx =13[t∫ f(t) dt ∫(∫ f(t)dt)dt]+c =13[tψ(t) ∫(∫ f(t)dt)dt]+c =13[x^3ψ(x^3) ∫ 3x^2ψ(x^3 ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

∫ f x dx =ψ x...

Question

$\int f (x) d x = ψ (x)$ , then $\int x^{5} f (x^{3}) d x$ is equal to

\frac{1}{3} [x^{3} ψ (x^{3}) - \int x^{2} ψ (x^{3}) d x] + c

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\frac{1}{3} x^{3} ψ (x^{3}) - \int x^{3} ψ (x^{3}) d x + c

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\frac{1}{3} x^{3} ψ (x^{3}) - \int x^{2} ψ (x^{3}) d x + c

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\frac{1}{3} [x^{3} ψ (x^{3}) - \int x^{3} ψ (x^{3}) d x] + c

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Solution

The correct option is C $\frac{1}{3} x^{3} ψ (x^{3}) - \int x^{2} ψ (x^{3}) d x + c$
Given, $\int f (x) d x = ψ (x)$
$I = \int x^{5} f (x^{3}) d x$
put $x^{3} = t \Rightarrow x^{2} d x = \frac{1}{3} d t$
$\Rightarrow I = \frac{1}{3} \int t f (t) d x$
$= \frac{1}{3} [t \int f (t) d t - \int (\int f (t) d t) d t] + c$
$= \frac{1}{3} [t ψ (t) - \int (\int f (t) d t) d t] + c$
$= \frac{1}{3} [x^{3} ψ (x^{3}) - \int 3 x^{2} ψ (x^{3}) d x] + c$
$= \frac{1}{3} x^{3} ψ (x^{3}) - \int x^{2} ψ (x^{3}) d x + c$

Suggest Corrections

∫f(x)dx=ψ(x), then ∫x5f(x3)dx is equal to

The correct option is C 13x3ψ(x3)−∫x2ψ(x3)dx+cGiven, ∫f(x)dx=ψ(x) I=∫x5f(x3)dx put x3=t⇒x2dx=13dt ⇒I=13∫tf(t)dx =13[t∫f(t)dt−∫(∫f(t)dt)dt]+c =13[tψ(t)−∫(∫f(t)dt)dt]+c =13[x3ψ(x3)−∫3x2ψ(x3)dx]+c =13x3ψ(x3)−∫x2ψ(x3)dx+c

$\int f (x) d x = ψ (x)$ , then $\int x^{5} f (x^{3}) d x$ is equal to