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Magazine Chapter 7: Problem 48
Evaluate the definite integral. \(\int_{0}^{\pi} x \sin x d x\)
Short Answer
Expert verified
The definite integral evaluates to \( \pi \).
Step by step solution
01
Integration by Parts Formula
To evaluate the integral \( \int_{0}^{\pi} x \sin x \, dx \), we will use integration by parts. The formula for integration by parts is given by: \[ \int u \, dv = uv - \int v \, du \] where \( u \) and \( dv \) are parts of the integrand chosen appropriately.
02
Identify Parts for Integration by Parts
In our integral \( \int x \sin x \, dx \), we will let:- \( u = x \), which means \( du = dx \)- \( dv = \sin x \, dx \), which gives \( v = \int \sin x \, dx = -\cos x \)
03
Apply Integration by Parts
Using the integration by parts formula, substitute \( u \), \( du \), \( v \), and \( dv \) into the equation:\[\int x \sin x \, dx = -x \cos x \bigg|_{0}^{\pi} + \int \cos x \, dx \bigg|_{0}^{\pi} \]
04
Evaluate Boundary Terms
Calculate the boundary terms for \(-x \cos x \bigg|_{0}^{\pi}\):- At \(x = \pi\), \(-x \cos x = -\pi \cos \pi = \pi\)- At \(x = 0\), \(-x \cos x = 0\)So the boundary term is \(\pi - 0 = \pi\).
05
Integrate Remaining Integral
Now, compute the integral \( \int \cos x \, dx \bigg|_{0}^{\pi} \):\[ \int \cos x \, dx = \sin x \bigg|_{0}^{\pi} \]Evaluate at the boundaries:- \( \sin \pi = 0 \)- \( \sin 0 = 0 \)Thus, \( \sin x \bigg|_{0}^{\pi} = 0 - 0 = 0 \).
06
Combine Results
Adding the results from Steps 4 and 5 gives the final value for the integral:\[ \pi + 0 = \pi \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is a helpful technique used to tackle integrals that involve products of functions, like \( \int x \sin x \, dx \). This method is derived from the product rule for differentiation, and the formula is given by \( \int u \, dv = uv - \int v \, du \). Here's how it works:
- First, we identify parts of the integrand \( x \sin x \) as \( u \) and \( dv \). In this case, we set \( u = x \) and \( dv = \sin x \, dx \).
- We then differentiate \( u \) to find \( du = dx \), and integrate \( dv \) to find \( v = -\cos x \).
Trigonometric Integrals
Trigonometric integrals involve integrating trigonometric functions, and they can sometimes be tricky. However, with the right approach, they are manageable. In our exercise, we deal with \( \sin x \) and \( \cos x \). Here are a few tips:
- The integral of \( \sin x \) is \( -\cos x \). This means, when integrating \( \sin x \, dx \), the result is \( -\cos x + C \).
- Similarly, the integral of \( \cos x \) is \( \sin x \). Hence, when we face an integral like \( \int \cos x \, dx \), it becomes \( \sin x + C \).
Boundary Evaluation
Boundary evaluation is crucial when dealing with definite integrals, like \( \int_{0}^{\pi} \). It involves evaluating the integrated function at the upper and lower bounds and subtracting the results.
- In our step 4, we evaluated \( -x \cos x |_{0}^{\pi} \). This means we find \( -\pi \cos \pi \) at the upper bound \( \pi \) and \( 0 \) at the lower bound \( 0 \), resulting in \( \pi \).
- The remaining integral was \( \int \cos x \ \, dx \). After integration, we use boundary evaluation again: \( \sin x |_{0}^{\pi} = 0 \), since both \( \sin 0 \) and \( \sin \pi \) are zero.
