IBP for definite integrals. We can also use integration by parts to calculate definite integrals if we combine our formula with Part 2 of the Fundamental Theorem of Calculus. If we do this, we get We can also rewrite this formula using the IBP mnemonic as int a ^ b f(x)g^ prime (x)dx=(f(x)g(x))| a ^ b – integrate g(x) * f’ * (x) dx from a to b . int a ^ b u dv=(uv)| a ^ b – integrate v du from a to b . Example. If we |et u=x ard dv=sin(x)dx( Rightarrow du = dxan * dv = – cos(x) rangle int 0 ^ pi x sin(x)dx=(-x cos(x))| 0 ^ pi – integrate (- cos(x)) dx from 0 to pi – (- x * cos(x)) | 0 ^ * -(-sin(x))| 0 ^ * – (- pi * cos(pi) + 0cos(0)) – (- sin(pi) + sin(0)) as in the previous video, then the formula above gives us
The Correct Answer and Explanation is :
To correctly solve the integral (\int_0^\pi x \sin(x) \, dx) using integration by parts (IBP), let’s follow the steps systematically:
Integration by Parts Formula:
[
\int_a^b u \, dv = \left[ u v \right]_a^b – \int_a^b v \, du
]
Step 1: Assign (u) and (dv)
Let:
[
u = x \quad \text{and} \quad dv = \sin(x) \, dx
]
Step 2: Compute (du) and (v)
[
du = dx \quad \text{and} \quad v = -\cos(x)
]
Step 3: Apply the Formula
Using the IBP formula:
[
\int_0^\pi x \sin(x) \, dx = \left[ -x \cos(x) \right]_0^\pi – \int_0^\pi (-\cos(x)) \, dx
]
Step 4: Evaluate Each Term
- Boundary term: (\left[ -x \cos(x) \right]_0^\pi)
[
= \left[ -\pi \cos(\pi) – (0 \cos(0)) \right]
= -\pi(-1) – 0(1) = \pi
] - Integral of (-\cos(x)):
[
\int_0^\pi -\cos(x) \, dx = -\left[ \sin(x) \right]_0^\pi
= -\left[ \sin(\pi) – \sin(0) \right]
= -[0 – 0] = 0
]
Step 5: Combine Results
[
\int_0^\pi x \sin(x) \, dx = \pi + 0 = \pi
]
Explanation (300 words)
Integration by parts (IBP) is a powerful technique to compute integrals of products of functions. It is derived from the product rule for derivatives and allows us to split an integral into two parts, making it easier to solve. The formula is:
[
\int u \, dv = uv – \int v \, du
]
In definite integrals, the limits of integration are applied to the boundary term, ensuring an accurate calculation.
In this example, we calculated (\int_0^\pi x \sin(x) \, dx) using IBP. Choosing (u = x) simplifies the problem because its derivative ((du = dx)) is simpler than the original function. We chose (dv = \sin(x) \, dx), knowing its integral is straightforward ((v = -\cos(x))). After substituting these into the IBP formula, the integral was broken into two parts: a boundary term, ([-x \cos(x)]_0^\pi), and a new integral, (\int_0^\pi -\cos(x) \, dx).
Evaluating the boundary term involved substituting the limits of integration for (x) and computing the cosine function. For the second term, we calculated the definite integral of (-\cos(x)), which simplifies due to the periodic nature of the sine function at the limits.
Adding these results gave the final answer, (\pi). This approach highlights the power of IBP in breaking down challenging integrals into manageable steps.