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Question: If \(\int {f(x)dx} = \Psi (x)\), then \(\int {{x^5}f\left( {{x^3}} \right)dx} \) is equal to: A. \...

If ∫f(x)dx=Ψ(x)\int {f(x)dx} = \Psi (x), then ∫x5f(x3)dx\int {{x^5}f\left( {{x^3}} \right)dx} is equal to:
A. 13x3Ψ(x3)−3∫x3Ψ(x3)dx+C\dfrac{1}{3}{x^3}\Psi \left( {{x^3}} \right) - 3\int {{x^3}\Psi \left( {{x^3}} \right)dx} + C
B. 13x3Ψ(x3)−∫x2Ψ(x3)dx+C\dfrac{1}{3}{x^3}\Psi \left( {{x^3}} \right) - \int {{x^2}\Psi \left( {{x^3}} \right)dx} + C
C. 13[x3Ψ(x3)−∫x3Ψ(x3)dx]+C\dfrac{1}{3}\left[ {{x^3}\Psi \left( {{x^3}} \right) - \int {{x^3}\Psi \left( {{x^3}} \right)dx} } \right] + C
D. 13[x3Ψ(x3)−∫x2Ψ(x3)dx]+C\dfrac{1}{3}\left[ {{x^3}\Psi \left( {{x^3}} \right) - \int {{x^2}\Psi \left( {{x^3}} \right)dx} } \right] + C

Explanation

Solution

Solve by integrating the given expression in parts: ∫f(x)g(x)dx=f(x)∫g(x)dx−∫[f′(x)∫g(x)dx]dx\int {f(x)g(x)dx} = f(x)\int {g(x)dx} - \int {\left[ {f'(x)\int {g(x)dx} } \right]dx} . Make an appropriate substitution. If we substitute x = f(t), then
dx = f′(t) dtdx{\text{ }} = {\text{ }}f'\left( t \right){\text{ }}dtand ∫f(x)dx=∫f[f(t)]f′(t)dt\int {f(x)dx} = \int {f\left[ {f(t)} \right]f'(t)dt} .
Substitute x3=t{x^3} = t and integrate by parts.
Recall that ddxxn=nxn−1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} and that if ∫f(x)dx=g(x)\int {f(x)dx} = g(x), then ∫f(y)dy=g(y)\int {f(y)dy} = g(y).

Complete step by step Solution:
Let's say I=∫x5f(x3)dxI = \int {{x^5}f\left( {{x^3}} \right)dx} , which can also be written as:
⇒I=∫x3.x2f(x3)dx\Rightarrow I = \int {{x^3}.{x^2}f\left( {{x^3}} \right)dx}

Substituting x3=t{x^3} = t, we have 3x2dx=dt3{x^2}dx = dt.
And, I=13∫tf(t)dtI = \dfrac{1}{3}\int {tf(t)dt} .

Integrating by parts, taking t as the first function and f(t) as the second function, we get:
⇒I=13[t∫f(t)dt−∫(ddtt∫f(t)dt)dt]+C\Rightarrow I = \dfrac{1}{3}\left[ {t\int {f(t)dt} - \int {\left( {\dfrac{d}{{dt}}t\int {f(t)dt} } \right)dt} } \right] + C

Differentiating ‘t’ w.r.t. ‘t’, we get ‘1’; so
⇒I=13[t∫f(t)dt−∫(∫f(t)dt)dt]+C\Rightarrow I = \dfrac{1}{3}\left[ {t\int {f(t)dt} - \int {\left( {\int {f(t)dt} } \right)dt} } \right] + C
It is given that ∫f(x)dx=Ψ(x)\int {f(x)dx} = \Psi (x)

On Substituting value in equation, we get;
⇒I=13[tΨ(t)−∫Ψ(t)dt]+C\Rightarrow I = \dfrac{1}{3}\left[ {t\Psi (t) - \int {\Psi (t)dt} } \right] + C
Back substituting x3=t{x^3} = t, we have 3x2dx=dt3{x^2}dx = dt, we get:

⇒I=13[x3Ψ(x3)−∫Ψ(x3)(3x2)dx]+C \Rightarrow I = \dfrac{1}{3}\left[ {{x^3}\Psi ({x^3}) - \int {\Psi ({x^3})\left( {3{x^2}} \right)dx} } \right] + C
⇒I=13x3Ψ(x3)−∫x2Ψ(x3)dx+C\Rightarrow I = \dfrac{1}{3}{x^3}\Psi ({x^3}) - \int {{x^2}\Psi ({x^3})dx} + C
Hence, the correct answer is B. 13x3Ψ(x3)−∫x2Ψ(x3)dx+C\dfrac{1}{3}{x^3}\Psi ({x^3}) - \int {{x^2}\Psi ({x^3})dx} + C.

Note: integration by parts or partial integration is a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative. When integrating by parts:

∫f(x)g(x)dx\int {f(x)g(x)dx} = f(x)∫g(x)dxf(x)\int {g(x)dx} − ∫[f′(x)∫g(x)dx]dx\int {\left[ {f'(x)\int {g(x)dx} } \right]dx} , make sure to select the functions f(x) and g(x) in such a manner that we can integrate the derivative of the first function f(x) easily. Usually, it's better to follow the rule of ILATE when selecting the first and the second functions.
I: Inverse Trigonometric Functions
L: Logarithmic Functions
A: Algebraic Functions
T: Trigonometric Functions
E: Exponential Functions

Application of Integrals are very important. Integrals are used to find the area under the curve and we can solve differential equations using integration.