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Chapter 5: Q 14. (page 428)

Provide a justification for each equality in the statement of the integration-by-parts formula for definite integrals from Theorem 5.10.

Short Answer

Expert verified

The definite integral of is a number and represents the area under the curve from a to b.

Step by step solution

01

Step 1. Given information

Theorem 5.10 is as follows:

If $u = u (x), v = v (x)$ are differentiable functions on $[a, b]$ , then $\int_{a}^{b} u d v = {[\int u d v]}_{a}^{b} = {[u v]}_{a}^{b} - \int_{a}^{b} v d u$

02

Step 2. Explanation

According to theorem 5.10,

$\int_{a}^{b} u d v = {[\int u d v]}_{a}^{b} = {[u v]}_{a}^{b} - \int_{a}^{b} v d u$

Here, $u (x), v (x)$ are differentiable functions on $[a, b]$

The definite integral of is a number and represents the area under the curve from a to b.

Take $u, d v$ as first and second functions respectively.

So, $\int_{a}^{b} u d v = {[\int u d v]}_{a}^{b} = {[u v]}_{a}^{b} - \int_{a}^{b} v d u$

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Most popular questions from this chapter

Why doesn’t the definite integral $\int_{2}^{3} \sqrt{1 - x^{2}} d x$ make sense? (Hint: Think about domains.)

Solve the integral: $\int (3 - x) e^{2 x} d x$

Solve given definite integral.

$\int_{- 1}^{1} x^{3} \sqrt{9 x^{2} - 1} d x$

Explain why $\int \frac{2 x}{x^{2} + 1} d x$ and $\int \frac{1}{x \ln x} d x$ are essentially the same integral after a change of variables.

Solve given definite integral.

$\int_{1}^{2} \frac{3}{{(9 - 2 x^{2})}^{3 / 2}} d x$

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