Evaluate / xe*dx using the integration-by-parts formula with u= x and dv = e2dx

Evaluate / xe*dx using the integration-by-parts formula with u= x and dv = e2dx. 3.2 Evaluate / xinx dx. 3.3 Evaluate / sinxdx.

The Correct Answer and Explanation is :

Let’s evaluate each of these integrals step-by-step, using appropriate methods like integration by parts or standard integration rules. I’ll walk you through the solution for each problem and explain the reasoning behind them.

3.1 Evaluate ∫ x * e^(2x) dx using Integration by Parts

We can solve this integral using the integration by parts formula:

[
\int u \, dv = uv – \int v \, du
]

Here, we can choose:

  • ( u = x ), so that ( du = dx )
  • ( dv = e^{2x} dx ), so we need to integrate ( e^{2x} ), which gives ( v = \frac{1}{2} e^{2x} )

Now, applying the integration by parts formula:

[
\int x e^{2x} dx = \left( x \cdot \frac{1}{2} e^{2x} \right) – \int \frac{1}{2} e^{2x} dx
]

Simplifying:

[
= \frac{1}{2} x e^{2x} – \frac{1}{2} \int e^{2x} dx
]

The integral of ( e^{2x} ) is ( \frac{1}{2} e^{2x} ), so:

[
= \frac{1}{2} x e^{2x} – \frac{1}{4} e^{2x} + C
]

Thus, the solution to the integral is:

[
\int x e^{2x} dx = \frac{1}{2} x e^{2x} – \frac{1}{4} e^{2x} + C
]

3.2 Evaluate ∫ x * sin(x) dx

For this integral, we will again use integration by parts. Let:

  • ( u = x ), so ( du = dx )
  • ( dv = \sin(x) dx ), so integrating gives ( v = -\cos(x) )

Applying the integration by parts formula:

[
\int x \sin(x) dx = \left( x \cdot -\cos(x) \right) – \int -\cos(x) dx
]

Simplifying:

[
= -x \cos(x) + \int \cos(x) dx
]

The integral of ( \cos(x) ) is ( \sin(x) ), so:

[
= -x \cos(x) + \sin(x) + C
]

Thus, the solution to the integral is:

[
\int x \sin(x) dx = -x \cos(x) + \sin(x) + C
]

3.3 Evaluate ∫ sin(x) dx

This is a simpler integral that doesn’t require integration by parts. The integral of ( \sin(x) ) is a standard result:

[
\int \sin(x) dx = -\cos(x) + C
]

Thus, the solution to the integral is:

[
\int \sin(x) dx = -\cos(x) + C
]

Explanation

  • Integration by Parts: This technique is useful when you have a product of two functions. The formula ( \int u \, dv = uv – \int v \, du ) helps break down the product into more manageable parts. By strategically choosing ( u ) and ( dv ), we simplify the integral and solve it step by step.
  • Basic Integrals: Some integrals, like ( \int \sin(x) dx ), are basic and do not require complex methods. These are commonly memorized as standard results.
  • Practice: Understanding the structure of different integrals—whether they are products of functions or simple trigonometric functions—will help you determine when to apply integration by parts and when a straightforward approach will suffice.

By following these methods and principles, we can evaluate a wide range of integrals.

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