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Antiderivative of x^3e^(2x)

Created by @nathanedwards
 at October 31st 2023, 5:12:28 pm.
AP_Calculus_AB-Questions
#calculus
#integration
#antiderivative

AP Calculus AB Exam Question:

Find the antiderivative of the function:

f(x)=x3e2x f(x) = x^3e^{2x}

Solution:

To find the antiderivative of the function f(x)=x3e2xf(x) = x^3e^{2x}, we will use integration by parts method.

Integration by parts formula states:

∫u dv=uv−∫v du \int u \, dv = uv - \int v \, du

where uu and vv are differentiable functions.

Let's choose:

u=x3⇒du=3x2 dx u = x^3 \quad \Rightarrow \quad du = 3x^2 \, dx
dv=e2x dx⇒v=12e2x dv = e^{2x} \, dx \quad \Rightarrow \quad v = \frac{1}{2}e^{2x}

Now, we can use the integration by parts formula to find the antiderivative:

∫x3e2x dx=12x3e2x−∫12e2x⋅3x2 dx=12x3e2x−32∫x2e2x dx \begin{align*} \int x^3e^{2x} \, dx &= \frac{1}{2}x^3e^{2x} - \int \frac{1}{2}e^{2x} \cdot 3x^2 \, dx \\ &= \frac{1}{2}x^3e^{2x} - \frac{3}{2} \int x^2e^{2x} \, dx \end{align*}

We now have another integral with the same form, but one degree lower. We will apply integration by parts again to evaluate it.

Choosing:

u=x2⇒du=2x dx u = x^2 \quad \Rightarrow \quad du = 2x \, dx
dv=e2x dx⇒v=12e2x dv = e^{2x} \, dx \quad \Rightarrow \quad v = \frac{1}{2}e^{2x}

we get:

∫x3e2x dx=12x3e2x−32(12x2e2x−∫12e2x⋅2x dx)=12x3e2x−32(12x2e2x−12∫e2x⋅2x dx)=12x3e2x−34x2e2x+34∫e2x⋅2x dx \begin{align*} \int x^3e^{2x} \, dx &= \frac{1}{2}x^3e^{2x} - \frac{3}{2} \left(\frac{1}{2}x^2e^{2x} - \int \frac{1}{2}e^{2x} \cdot 2x \, dx\right) \\ &= \frac{1}{2}x^3e^{2x} - \frac{3}{2}\left(\frac{1}{2}x^2e^{2x} - \frac{1}{2} \int e^{2x} \cdot 2x \, dx\right) \\ &= \frac{1}{2}x^3e^{2x} - \frac{3}{4}x^2e^{2x} + \frac{3}{4} \int e^{2x} \cdot 2x \, dx \\ \end{align*}

At this point, we have a new integral:

∫e2x⋅2x dx \int e^{2x} \cdot 2x \, dx

We can apply integration by parts once more:

Choosing:

u=x⇒du=dx u = x \quad \Rightarrow \quad du = dx
dv=e2x dx⇒v=12e2x dv = e^{2x} \, dx \quad \Rightarrow \quad v = \frac{1}{2}e^{2x}

we get:

12x3e2x−34x2e2x+34∫e2x⋅2x dx=12x3e2x−34x2e2x+34(12x⋅e2x−∫12e2x dx)=12x3e2x−34x2e2x+34(12x⋅e2x−12∫e2x dx) \begin{align*} \frac{1}{2}x^3e^{2x} - \frac{3}{4}x^2e^{2x} + \frac{3}{4} \int e^{2x} \cdot 2x \, dx &= \frac{1}{2}x^3e^{2x} - \frac{3}{4}x^2e^{2x} + \frac{3}{4} \left(\frac{1}{2}x\cdot e^{2x} - \int \frac{1}{2}e^{2x} \, dx\right) \\ &= \frac{1}{2}x^3e^{2x} - \frac{3}{4}x^2e^{2x} + \frac{3}{4} \left(\frac{1}{2}x\cdot e^{2x} - \frac{1}{2}\int e^{2x} \, dx\right) \\ \end{align*}

The integral remaining, ∫e2x dx\int e^{2x} \, dx, is a simple integration:

∫e2x dx=12e2x \int e^{2x} \, dx = \frac{1}{2}e^{2x}

Substituting this back into the previous equation, we get:

12x3e2x−34x2e2x+14x⋅e2x−38e2x+C \begin{align*} \frac{1}{2}x^3e^{2x} - \frac{3}{4}x^2e^{2x} + \frac{1}{4}x\cdot e^{2x} - \frac{3}{8}e^{2x} + C \end{align*}

Therefore, the antiderivative of f(x)=x3e2xf(x) = x^3e^{2x} is:

F(x)=12x3e2x−34x2e2x+14x⋅e2x−38e2x+C F(x) = \frac{1}{2}x^3e^{2x} - \frac{3}{4}x^2e^{2x} + \frac{1}{4}x\cdot e^{2x} - \frac{3}{8}e^{2x} + C

where CC is the constant of integration.

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