calculus 2 integral practice Solved: calculus 2

Hey there! Today we're going to tackle some practice problems for indefinite integrals using integration by parts. This can be a tricky concept to master, but with practice and determination, you'll get the hang of it in no time!

Practice Problem 1:

Find the indefinite integral of x²sin(x).

To solve this problem, we'll use integration by parts. Let u = x² and dv/dx = sin(x). Then:

• du/dx = 2x
• v = -cos(x)

Using the formula for integration by parts, we have:

∫x²sin(x)dx = -x²cos(x) + 2xsin(x) - 2∫xsin(x)dx

We now have another integral to solve, but this time we'll use integration by parts with u = x and dv/dx = sin(x). Then:

• du/dx = 1
• v = -cos(x)

Using the formula again, we have:

∫xsin(x)dx = -xcos(x) + ∫cos(x)dx

This last integral is easy to solve: ∫cos(x)dx = sin(x). So, our final answer is:

∫x²sin(x)dx = -x²cos(x) + 2xsin(x) + 2xcos(x) - 2sin(x) + C

Practice Problem 2:

Find the indefinite integral of (x + 1)²e^2x.

This problem is a bit more complicated, but we can still use integration by parts. Let u = (x + 1)² and dv/dx = e^2x. Then:

• du/dx = 2(x + 1)
• v = 1/2 e^2x

Using the formula, we have:

∫(x + 1)²e^2xdx = 1/2 (x + 1)²e^2x - 2∫(x + 1)e^2xdx

For the next step, we'll use integration by parts again. Let u = x + 1 and dv/dx = e^2x. Then:

• du/dx = 1
• v = 1/2 e^2x

Using the formula once more, we have:

∫(x + 1)e^2xdx = 1/2 (x + 1)e^2x - 1/4 e^2x + C

Now we can substitute this result back into our original formula and simplify:

∫(x + 1)²e^2xdx = 1/2 (x + 1)²e^2x - (x + 1)e^2x + 1/2 e^2x + C

Well, that's it for today's practice problems! I hope this has helped you feel more confident with integration by parts for indefinite integrals. Keep practicing and you'll be an expert in no time!

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