Express the integral as a limit of Riemann sums. Do not evaluate the limit.
Transcribed Image Text:**Problem 26: Evaluate the integral**
\[
\int_{0}^{2\pi} x^2 \sin x \, dx
\]
This integral involves finding the area under the curve of the function \( x^2 \sin x \) from \( x = 0 \) to \( x = 2\pi \).
To solve this, you can use integration by parts. Integration by parts is based on the formula:
\[
\int u \, dv = uv - \int v \, du
\]
Choose \( u \) and \( dv \) such that \( du \) and \( v \) are easy to compute.
Let:
\[
u = x^2 \quad \Rightarrow \quad du = 2x \, dx
\]
\[
dv = \sin x \, dx \quad \Rightarrow \quad v = -\cos x
\]
Using the integration by parts formula:
\[
\int_{0}^{2\pi} x^2 \sin x \, dx = \left. -x^2 \cos x \right|_{0}^{2\pi} + \int_{0}^{2\pi} 2x \cos x \, dx
\]
Evaluate \(\left. -x^2 \cos x \right|_{0}^{2\pi}\):
\[
-x^2 \cos x \Bigg|_{0}^{2\pi} = \left. -x^2 \cos x \right|_{2\pi} + \left. x^2 \cos x \right|_{0} = -(2\pi)^2 \cos(2\pi) + 0^2 \cos(0) = -4\pi^2(1) + 0 = -4\pi^2
\]
Now, we need to evaluate \( \int_{0}^{2\pi} 2x \cos x \, dx \). Again using integration by parts, let:
\[
u = 2x \quad \Rightarrow \quad du = 2 \, dx
\]
\[
dv = \cos x \, dx \quad \Rightarrow \quad v = \sin x
\]
Now,
\[
\int_{0}^{2\pi} 2x \cos x \, dx = \left. 2x \
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
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Transcribed Image Text:### Educational Instructions: Understanding Riemann Sums
**Problem 25-26:**
Express the integral as a limit of Riemann sums. Do not evaluate the limit.
---
**Explanation:**
In calculus, one fundamental technique to approximate the value of definite integrals is through the concept of Riemann sums. This problem requires you to express the integral function as a limit of Riemann sums without evaluating the limit.
Here is a step-by-step approach to understanding this concept:
1. **Partition the Interval:**
Divide the interval \([a, b]\) into \(n\) subintervals of equal width \(\Delta x = \frac{b-a}{n}\).
2. **Choose Sample Points:**
For each subinterval, choose a sample point \(x_i^*\). This point can be the left endpoint, right endpoint, or midpoint of the subinterval. For simplicity, assuming \(x_i^*\) is the right endpoint, then \(x_i^* = a + i \Delta x\).
3. **Form the Riemann Sum:**
The Riemann Sum \(S_n\) for the function \(f(x)\) over the interval \([a, b]\) is given by the sum:
\[
S_n = \sum_{i=1}^{n} f(x_i^*) \Delta x
\]
4. **Take the Limit:**
To express the integral as a limit, take the limit of the Riemann sum as the number of subintervals \(n\) approaches infinity:
\[
\int_{a}^{b} f(x) dx = \lim_{{n \to \infty}} \sum_{i=1}^{n} f(x_i^*) \Delta x
\]
By following these steps, you can express a definite integral as the limit of Riemann sums, which is essential in forming the foundation for more advanced integral calculus.
Remember, in this activity, you do not need to evaluate the limit—just express the integral in the form given above.
Transcribed Image Text:**Problem 26.**
Evaluate the integral:
\[
\int_{0}^{2\pi} x^2 \sin x \, dx
\]
This integral requires the use of integration techniques such as integration by parts.
Integration by parts is based on the product rule for differentiation and is given by the formula:
\[
\int u \, dv = uv - \int v \, du
\]
For this integral, we need to choose \( u \) and \( dv \) appropriately. Let's consider \( u = x^2 \) and \( dv = \sin x \, dx \).
Then we compute \( du \) and \( v \):
\[
du = 2x \, dx
\]
\[
v = -\cos x
\]
Now we apply the integration by parts formula:
\[
\int_{0}^{2\pi} x^2 \sin x \, dx = \left. x^2 (- \cos x) \right|_0^{2\pi} - \int_{0}^{2\pi} (- \cos x) (2x) \, dx
\]
Evaluating the first term:
\[
\left. x^2 (- \cos x) \right|_0^{2\pi} = (- (2\pi)^2 \cos (2\pi)) - (0^2 (- \cos 0)) = -4\pi^2 \cdot 1 = -4\pi^2
\]
For the second term, we substitute it back into another integration by parts:
\[
\int_{0}^{2\pi} 2x \cos x \, dx
\]
Set \( u = 2x \) and \( dv = \cos x \, dx \).
Compute \( du \) and \( v \):
\[
du = 2 \, dx
\]
\[
v = \sin x
\]
Apply integration by parts again:
\[
\int_{0}^{2\pi} 2x \cos x \, dx = \left. 2x \sin x \right|_0^{2\pi} - \int_{0}^{2\pi} 2 \sin x \, dx
\]
Evaluate the first term:
\[
\left. 2x \sin x \right|_0^{2\pi
![**Problem 26: Evaluate the integral**
\[
\int_{0}^{2\pi} x^2 \sin x \, dx
\]
This integral involves finding the area under the curve of the function \( x^2 \sin x \) from \( x = 0 \) to \( x = 2\pi \).
To solve this, you can use integration by parts. Integration by parts is based on the formula:
\[
\int u \, dv = uv - \int v \, du
\]
Choose \( u \) and \( dv \) such that \( du \) and \( v \) are easy to compute.
Let:
\[
u = x^2 \quad \Rightarrow \quad du = 2x \, dx
\]
\[
dv = \sin x \, dx \quad \Rightarrow \quad v = -\cos x
\]
Using the integration by parts formula:
\[
\int_{0}^{2\pi} x^2 \sin x \, dx = \left. -x^2 \cos x \right|_{0}^{2\pi} + \int_{0}^{2\pi} 2x \cos x \, dx
\]
Evaluate \(\left. -x^2 \cos x \right|_{0}^{2\pi}\):
\[
-x^2 \cos x \Bigg|_{0}^{2\pi} = \left. -x^2 \cos x \right|_{2\pi} + \left. x^2 \cos x \right|_{0} = -(2\pi)^2 \cos(2\pi) + 0^2 \cos(0) = -4\pi^2(1) + 0 = -4\pi^2
\]
Now, we need to evaluate \( \int_{0}^{2\pi} 2x \cos x \, dx \). Again using integration by parts, let:
\[
u = 2x \quad \Rightarrow \quad du = 2 \, dx
\]
\[
dv = \cos x \, dx \quad \Rightarrow \quad v = \sin x
\]
Now,
\[
\int_{0}^{2\pi} 2x \cos x \, dx = \left. 2x \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4e699669-58d4-4b70-bb83-6285c2150cf2%2F9a6fd2f4-7d3c-42d1-a782-06b25a543ed7%2Fpg1e8du_processed.png&w=3840&q=75)
![### Educational Instructions: Understanding Riemann Sums
**Problem 25-26:**
Express the integral as a limit of Riemann sums. Do not evaluate the limit.
---
**Explanation:**
In calculus, one fundamental technique to approximate the value of definite integrals is through the concept of Riemann sums. This problem requires you to express the integral function as a limit of Riemann sums without evaluating the limit.
Here is a step-by-step approach to understanding this concept:
1. **Partition the Interval:**
Divide the interval \([a, b]\) into \(n\) subintervals of equal width \(\Delta x = \frac{b-a}{n}\).
2. **Choose Sample Points:**
For each subinterval, choose a sample point \(x_i^*\). This point can be the left endpoint, right endpoint, or midpoint of the subinterval. For simplicity, assuming \(x_i^*\) is the right endpoint, then \(x_i^* = a + i \Delta x\).
3. **Form the Riemann Sum:**
The Riemann Sum \(S_n\) for the function \(f(x)\) over the interval \([a, b]\) is given by the sum:
\[
S_n = \sum_{i=1}^{n} f(x_i^*) \Delta x
\]
4. **Take the Limit:**
To express the integral as a limit, take the limit of the Riemann sum as the number of subintervals \(n\) approaches infinity:
\[
\int_{a}^{b} f(x) dx = \lim_{{n \to \infty}} \sum_{i=1}^{n} f(x_i^*) \Delta x
\]
By following these steps, you can express a definite integral as the limit of Riemann sums, which is essential in forming the foundation for more advanced integral calculus.
Remember, in this activity, you do not need to evaluate the limit—just express the integral in the form given above.](https://content.bartleby.com/qna-images/question/4e699669-58d4-4b70-bb83-6285c2150cf2/80c56651-fbee-4abb-8f27-ce41e078caf2/s5ruwqh_processed.png)
![**Problem 26.**
Evaluate the integral:
\[
\int_{0}^{2\pi} x^2 \sin x \, dx
\]
This integral requires the use of integration techniques such as integration by parts.
Integration by parts is based on the product rule for differentiation and is given by the formula:
\[
\int u \, dv = uv - \int v \, du
\]
For this integral, we need to choose \( u \) and \( dv \) appropriately. Let's consider \( u = x^2 \) and \( dv = \sin x \, dx \).
Then we compute \( du \) and \( v \):
\[
du = 2x \, dx
\]
\[
v = -\cos x
\]
Now we apply the integration by parts formula:
\[
\int_{0}^{2\pi} x^2 \sin x \, dx = \left. x^2 (- \cos x) \right|_0^{2\pi} - \int_{0}^{2\pi} (- \cos x) (2x) \, dx
\]
Evaluating the first term:
\[
\left. x^2 (- \cos x) \right|_0^{2\pi} = (- (2\pi)^2 \cos (2\pi)) - (0^2 (- \cos 0)) = -4\pi^2 \cdot 1 = -4\pi^2
\]
For the second term, we substitute it back into another integration by parts:
\[
\int_{0}^{2\pi} 2x \cos x \, dx
\]
Set \( u = 2x \) and \( dv = \cos x \, dx \).
Compute \( du \) and \( v \):
\[
du = 2 \, dx
\]
\[
v = \sin x
\]
Apply integration by parts again:
\[
\int_{0}^{2\pi} 2x \cos x \, dx = \left. 2x \sin x \right|_0^{2\pi} - \int_{0}^{2\pi} 2 \sin x \, dx
\]
Evaluate the first term:
\[
\left. 2x \sin x \right|_0^{2\pi](https://content.bartleby.com/qna-images/question/4e699669-58d4-4b70-bb83-6285c2150cf2/80c56651-fbee-4abb-8f27-ce41e078caf2/27t0xy5_processed.png)
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