If ∫ limits f(x) dx = ψ (x) , then ∫ limits x5 f (x3) dx is equal to

Question

If ∫ limits f(x) dx = ψ (x) , then ∫ limits x5 f (x3) dx is equal to (A)  (1/3)[x3 ψ (x3)- ∫ limits x2 ψ (x3)dx]+c (B)  (1/3)x3 ψ (x3)-3 ∫ limits x3 ψ (x3)dx+c (C)  (1/3)x3 ψ (x3)- ∫ limits x2 ψ (x3)dx+c (D)  (1/3)[x3 ψ (x3)- ∫ limits x3 ψ (x3)dx]+c

Tardigrade Admin · Accepted Answer

Given, ∫ f(x) d x=ψ(x) Let I=∫ x5 f(x3) d x Put x3=t ⇒ x2 d x=(d t/3) .....(i) ∴ I=(1/3) ∫ t f(t) d t =(1/3)[t ⋅ ∫ f(t) d t-∫ (d/d t)(t) ∫ f(t) d t d t] [integration by parts] =(1/3)[t ψ(t)-∫ ψ(t) d t] =(1/3)[x3 ψ(x3)-3 ∫ x2 ψ(x3) d x]+c[from Eq. (i)] =(1/3) x3 ψ(x3)-∫ x2 ψ(x3) d x+c

Q. If $\int f (x) d x = ψ (x), t h e n \int x^{5} f (x^{3}) d x$ is equal to

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

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

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

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

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

31​[x3ψ(x3)−∫x2ψ(x3)dx]+c

31​x3ψ(x3)−3∫x3ψ(x3)dx+c

31​x3ψ(x3)−∫x2ψ(x3)dx+c

31​[x3ψ(x3)−∫x3ψ(x3)dx]+c

Given, ∫f(x)dx=ψ(x) Let I=∫x5f(x3)dx Put x3=t ⇒x2dx=3dt​.....(i) ∴I=31​∫tf(t)dt =31​[t⋅∫f(t)dt−∫{dtd​(t)∫f(t)dt}dt] [integration by parts] =31​[tψ(t)−∫ψ(t)dt] =31​[x3ψ(x3)−3∫x2ψ(x3)dx]+c[from Eq. (i)] =31​x3ψ(x3)−∫x2ψ(x3)dx+c

Q. If $\int f (x) d x = ψ (x), t h e n \int x^{5} f (x^{3}) d x$ is equal to

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

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

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

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