Antiderivative of f(x) = x · e²ˣ

This problem asks to find the antiderivative F(x) of the function f(x) = x · e²ˣ satisfying the condition F(0) = 0.

Problem

$f (x) = x \cdot e^{2 x}, F (0) = 0$

Solution

Step 1

Apply the integration by parts formula:

\int u d v = u v - \int v d u

Step 2

Choose u = x (then du = dx) and dv = e²ˣ dx (then v = e²ˣ/2):

u = x, d u = d x

d v = e^{2 x} d x, v = \frac{1}{2} e^{2 x}

Step 3

Substitute into the integration by parts formula:

\int x e^{2 x} d x = \frac{x e^{2 x}}{2} - \int \frac{e^{2 x}}{2} d x

Step 4

Evaluate the remaining integral and obtain the general expression for the antiderivative:

F (x) = \frac{x e^{2 x}}{2} - \frac{e^{2 x}}{4} + C

Step 5

Find the constant C from the condition F(0) = 0. Substitute x = 0: e⁰ = 1, therefore:

F (0) = 0 - \frac{1}{4} + C = 0

Step 6

From this, C = 1/4. The required antiderivative has the form:

Answer

$F (x) = \frac{x e^{2 x}}{2} - \frac{e^{2 x}}{4} + \frac{1}{4}$

Frequently asked questions

How do I find the antiderivative of f(x) = x · e²ˣ?

Apply integration by parts ∫ u dv = uv − ∫ v du. Choose u = x (the polynomial is reduced when differentiated) and dv = e²ˣ dx (easily integrated to e²ˣ/2).

Why is u = x chosen in this problem?

In integration by parts, u is usually taken to be the polynomial (here x) and dv to be the exponential. After differentiation, x turns into 1, and the remaining integral becomes tabular: ∫ e²ˣ dx.

Why is v = e²ˣ/2 and not e²ˣ?

Because integrating ∫ e²ˣ dx gives e²ˣ/2. Check: the derivative of e²ˣ/2 is e²ˣ · 2/2 = e²ˣ. The factor 1/2 appears due to the coefficient of x in the exponent.

When is integration by parts used?

When the integrand is a product of different types of functions: polynomial × exponential, polynomial × trigonometric, exponential × trigonometric, logarithm, or inverse trigonometric. The classic sign is that the integral is not taken directly but simplifies after applying the formula ∫ u dv = uv − ∫ v du.