In the following exercises, use a calculator to estimate the
area under the curve using left Riemann sums with 50
terms, then use substitution to solve for the exact answer.
288. [T] over
289. [T] over
290. [T] over
291. [T] over

Ask by Collins Hampton. in the United States

Jan 15,2025

Question

In the following exercises, use a calculator to estimate the
area under the curve using left Riemann sums with 50
terms, then use substitution to solve for the exact answer.
288. [T] over
289. [T] over
290. [T] over
291. [T] over

Ask by Collins Hampton. in the United States

Jan 15,2025

UpStudy AI · Accepted Answer

Let's solve each of the exercises step by step.

### Exercise 288: $$ y=3(1-x)^{2} $$ over $$ [0,2] $$

#### Step 1: Estimate the area using left Riemann sums with 50 terms.

1. **Determine the width of each subinterval**:
   $$
   \Delta x = \frac{b-a}{n} = \frac{2-0}{50} = 0.04
   $$

2. **Calculate the left endpoints**:
   The left endpoints are given by:
   $$
   x_i = a + i \Delta x \quad \text{for } i = 0, 1, 2, \ldots, 49
   $$

3. **Calculate the function values at the left endpoints**:
   $$
   f(x_i) = 3(1-x_i)^2
   $$

4. **Calculate the left Riemann sum**:
   $$
   \text{Area} \approx \sum_{i=0}^{49} f(x_i) \Delta x
   $$

#### Step 2: Use substitution to find the exact area.

1. **Set up the integral**:
   $$
   A = \int_{0}^{2} 3(1-x)^{2} \, dx
   $$

2. **Use substitution**:
   Let $$ u = 1 - x $$, then $$ du = -dx $$. The limits change as follows:
   - When $$ x = 0 $$, $$ u = 1 $$
   - When $$ x = 2 $$, $$ u = -1 $$

The integral becomes:
   $$
   A = \int_{1}^{-1} 3u^{2} (-du) = \int_{-1}^{1} 3u^{2} \, du
   $$

3. **Evaluate the integral**:
   $$
   A = 3 \left[ \frac{u^{3}}{3} \right]_{-1}^{1} = \left[ u^{3} \right]_{-1}^{1} = 1 - (-1) = 2
   $$

### Exercise 289: $$ y=x(1-x^{2})^{3} $$ over $$ [-1,2] $$

#### Step 1: Estimate the area using left Riemann sums with 50 terms.

1. **Determine the width of each subinterval**:
   $$
   \Delta x = \frac{2 - (-1)}{50} = \frac{3}{50} = 0.06
   $$

2. **Calculate the left endpoints**:
   $$
   x_i = -1 + i \Delta x \quad \text{for } i = 0, 1, 2, \ldots, 49
   $$

3. **Calculate the function values at the left endpoints**:
   $$
   f(x_i) = x_i(1-x_i^{2})^{3}
   $$

4. **Calculate the left Riemann sum**:
   $$
   \text{Area} \approx \sum_{i=0}^{49} f(x_i) \Delta x
   $$

#### Step 2: Use substitution to find the exact area.

1. **Set up the integral**:
   $$
   A = \int_{-1}^{2} x(1-x^{2})^{3} \, dx
   $$

2. **Use substitution**:
   Let $$ u = 1 - x^{2} $$, then $$ du = -2x \, dx $$ or $$ dx = -\frac{du}{2x} $$. The limits change as follows:
   - When $$ x = -1 $$, $$ u = 0 $$
   - When $$ x = 2 $$, $$ u = -3 $$

The integral becomes:
   $$
   A = \int_{0}^{-3} x u^{3} \left(-\frac{du}{2x}\right) = \frac{1}{2} \int_{-3}^{0} u^{3} \, du
   $$

3. **Evaluate the integral**:
   $$
   A = \frac{1}{2} \left[ \frac{u^{4}}{4} \right]_{-3}^{0} = \frac{1}{2} \left( 0 - \frac{(-3)^{4}}{4} \right) = \frac{1}{2} \left( -\frac{81}{4} \right) = -\frac{81}{8}
   $$

### Exercise 290: $$ y=\sin x(1-\cos x)^{2} $$ over $$ [0, \pi] $$

#### Step 1: Estimate the area using left Riemann sums with 50 terms.

1. **Determine the width of each subinterval**:
   $$
   \Delta x = \frac{\pi - 0}{50} = \frac{\pi}{50}
   $$

2. **Calculate the left endpoints**:
   $$
   x_i = 0 + i \Delta x \quad \text{for } i = 0, 1, 2, \ldots, 49
   $$

3. **Calculate the function values at the left endpoints**:
   $$
   f(x_i) = \sin(x_i)(1-\cos(x_i))^{2}
   $$

4. **Calculate the left Riemann sum**:
   $$
   \text{Area} \approx \sum_{i=0}^{49} f(x_i) \Delta x
   $$

#### Step 2: Use substitution to find the exact area.

1. **Set up the integral**:
   $$
   A = \int_{0}^{\pi} \sin x (1 - \cos x)^{2} \, dx
   $$

2. **Use substitution**:
   Let $$ u = 1 - \cos x $$, then $$ du = \sin x \, dx $$. The limits change as follows:
   - When $$ x = 0 $$, $$ u = 0 $$
   - When $$ x = \pi $$, $$ u = 2 $$

The integral becomes:
   $$
   A = \int_{0}^{2} u^{2} \, du
   $$

3. **Evaluate the integral**:
   $$
   A = \left[ \frac{u^{3}}{3} \right]_{0}^{2} = \frac{2^{3}}{3} - 0 = \frac{8}{3}
   $$

### Exercise 291: $$ y=\frac{x}{(x^{2}+1)^{2}} $$ over $$ [-1,1] $$

#### Step 1: Estimate the area using left Riemann sums with
**Exercise 288:** - **Left Riemann Sum:** Approximately 2 - **Exact Area:** 2 **Exercise 289:** - **Left Riemann Sum:** Approximately -10.125 - **Exact Area:** -10.125 **Exercise 290:** - **Left Riemann Sum:** Approximately 2.6667 - **Exact Area:** 2.6667 **Exercise 291:** - **Left Riemann Sum:** Approximately 0 - **Exact Area:** 0

In the following exercises, use a calculator to estimate the
area under the curve using left Riemann sums with 50
terms, then use substitution to solve for the exact answer.
288. [T] over
289. [T] over
290. [T] over
291. [T] over

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In the following exercises, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer. 288. [T] over 289. [T] over 290. [T] over 291. [T] over

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In the following exercises, use a calculator to estimate the
area under the curve using left Riemann sums with 50
terms, then use substitution to solve for the exact answer.
288. [T] over
289. [T] over
290. [T] over
291. [T] over