Solveeit Logo
Login

Question

Question: If $\int x^5 e^{-4x^3} dx = \frac{1}{48}e^{-4x^3} f(x) + C$, where C is a constant of integration, t...

If ∫x5e−4x3dx=148e−4x3f(x)+C\int x^5 e^{-4x^3} dx = \frac{1}{48}e^{-4x^3} f(x) + C, where C is a constant of integration, then f(x) is equal to

A

4x3+14x^3 + 1

B

−4x3−1-4x^3 - 1

C

−2x3−1-2x^3 - 1

D

−2x3+1-2x^3 + 1

Answer

-4x^3 - 1

Explanation

Solution

We can solve this problem using two methods: direct integration or differentiation.

Method 1: Direct Integration

Let's evaluate the integral ∫x5e−4x3dx\int x^5 e^{-4x^3} dx. This integral can be simplified using a substitution. Let t=−4x3t = -4x^3.

Then, differentiate tt with respect to xx:

dt=−12x2dxdt = -12x^2 dx

From this, we can express x2dxx^2 dx as:

x2dx=−112dtx^2 dx = -\frac{1}{12} dt

Also, from the substitution t=−4x3t = -4x^3, we have x3=−t4x^3 = -\frac{t}{4}.

Now, rewrite the integral by splitting x5x^5 into x3⋅x2x^3 \cdot x^2:

∫x5e−4x3dx=∫x3⋅e−4x3⋅x2dx\int x^5 e^{-4x^3} dx = \int x^3 \cdot e^{-4x^3} \cdot x^2 dx

Substitute x3=−t4x^3 = -\frac{t}{4}, e−4x3=ete^{-4x^3} = e^t, and x2dx=−112dtx^2 dx = -\frac{1}{12} dt:

∫(−t4)et(−112dt)=∫t48etdt=148∫tetdt\int \left(-\frac{t}{4}\right) e^t \left(-\frac{1}{12} dt\right) = \int \frac{t}{48} e^t dt = \frac{1}{48} \int t e^t dt

Now, we need to evaluate the integral ∫tetdt\int t e^t dt using integration by parts. The formula for integration by parts is ∫udv=uv−∫vdu\int u dv = uv - \int v du.

Let u=tu = t and dv=etdtdv = e^t dt. Then, du=dtdu = dt and v=etv = e^t. Applying the formula:

∫tetdt=tet−∫etdt=tet−et=et(t−1)\int t e^t dt = t e^t - \int e^t dt = t e^t - e^t = e^t(t - 1)

Now, substitute this result back into our main integral:

∫x5e−4x3dx=148et(t−1)+C\int x^5 e^{-4x^3} dx = \frac{1}{48} e^t (t - 1) + C

Finally, substitute t=−4x3t = -4x^3 back into the expression:

∫x5e−4x3dx=148e−4x3(−4x3−1)+C\int x^5 e^{-4x^3} dx = \frac{1}{48} e^{-4x^3} (-4x^3 - 1) + C

Comparing this result with the given form 148e−4x3f(x)+C\frac{1}{48}e^{-4x^3} f(x) + C, we can identify f(x)f(x):

f(x)=−4x3−1f(x) = -4x^3 - 1

Method 2: Differentiation

Given the equation:

∫x5e−4x3dx=148e−4x3f(x)+C\int x^5 e^{-4x^3} dx = \frac{1}{48}e^{-4x^3} f(x) + C

Differentiate both sides with respect to xx. The derivative of an integral is the integrand itself:

x5e−4x3=ddx[148e−4x3f(x)]x^5 e^{-4x^3} = \frac{d}{dx} \left[ \frac{1}{48}e^{-4x^3} f(x) \right]

Apply the product rule for differentiation, ddx(uv)=u′v+uv′\frac{d}{dx}(uv) = u'v + uv', where u=148e−4x3u = \frac{1}{48}e^{-4x^3} and v=f(x)v = f(x).

First, find the derivative of uu:

ddx(148e−4x3)=148e−4x3⋅ddx(−4x3)=148e−4x3(−12x2)=−1248x2e−4x3=−14x2e−4x3\frac{d}{dx}\left(\frac{1}{48}e^{-4x^3}\right) = \frac{1}{48} e^{-4x^3} \cdot \frac{d}{dx}(-4x^3) = \frac{1}{48} e^{-4x^3} (-12x^2) = -\frac{12}{48} x^2 e^{-4x^3} = -\frac{1}{4} x^2 e^{-4x^3}

Now apply the product rule:

x5e−4x3=(−14x2e−4x3)f(x)+148e−4x3f′(x)x^5 e^{-4x^3} = \left(-\frac{1}{4} x^2 e^{-4x^3}\right) f(x) + \frac{1}{48} e^{-4x^3} f'(x)

Divide the entire equation by e−4x3e^{-4x^3} (since e−4x3≠0e^{-4x^3} \neq 0):

x5=−14x2f(x)+148f′(x)x^5 = -\frac{1}{4} x^2 f(x) + \frac{1}{48} f'(x)

Multiply the entire equation by 48 to clear the denominators:

48x5=−12x2f(x)+f′(x)48x^5 = -12x^2 f(x) + f'(x)

Now, we can test the given options for f(x)f(x).

Let's test option (2): f(x)=−4x3−1f(x) = -4x^3 - 1. If f(x)=−4x3−1f(x) = -4x^3 - 1, then f′(x)=ddx(−4x3−1)=−12x2f'(x) = \frac{d}{dx}(-4x^3 - 1) = -12x^2.

Substitute f(x)f(x) and f′(x)f'(x) into the equation:

48x5=−12x2(−4x3−1)+(−12x2)=(48x5+12x2)−12x2=48x548x^5 = -12x^2 (-4x^3 - 1) + (-12x^2) = (48x^5 + 12x^2) - 12x^2 = 48x^5

Since both sides are equal, f(x)=−4x3−1f(x) = -4x^3 - 1 is the correct function.