R is the region bounded by the functions f(x)=3e^x−1 and g(x)=x^2−1. Find the area of the region bounded by the functions on the interval [−1,1].

To find the area of the region bounded by the two functions \( f(x) = 3e^x - 1 \) and \( g(x) = x^2 - 1 \) on the interval \([-1, 1]\), we follow these steps:

### Step 1: Set up the integral
We need to find the points where \( f(x) = g(x) \), which will help us identify the regions where one function is above the other.

So, we solve for the intersection points by setting:
\[
3e^x - 1 = x^2 - 1
\]
This simplifies to:
\[
3e^x = x^2
\]

This equation doesn't have a simple analytical solution, so we check the interval \([-1, 1]\) to see where the functions intersect by evaluating \( f(x) \) and \( g(x) \) at critical points.

### Step 2: Evaluate at critical points
At \( x = -1 \):
\[
f(-1) = 3e^{-1} - 1 \approx 3(0.3679) - 1 \approx 1.1037 - 1 = 0.1037
\]
\[
g(-1) = (-1)^2 - 1 = 0
\]

At \( x = 1 \):
\[
f(1) = 3e^{1} - 1 \approx 3(2.718) - 1 = 8.154 - 1 = 7.154
\]
\[
g(1) = 1^2 - 1 = 0
\]

At \( x = 0 \):
\[
f(0) = 3e^{0} - 1 = 3(1) - 1 = 2
\]
\[
g(0) = 0^2 - 1 = -1
\]

Based on these values, \( f(x) \) is above \( g(x) \) over the entire interval \([-1, 1]\).

### Step 3: Set up the integral for the area
The area between two curves is given by the integral of the difference between the functions:
\[
\text{Area} = \int_{-1}^{1} (f(x) - g(x)) \, dx
\]

Thus, we need to compute:
\[
\text{Area} = \int_{-1}^{1} \left( (3e^x - 1) - (x^2 - 1) \right) dx
\]
Simplifying the integrand:
\[
\text{Area} = \int_{-1}^{1} \left( 3e^x - x^2 \right) dx
\]

### Step 4: Compute the integral
We now split the integral into two parts:
\[
\text{Area} = \int_{-1}^{1} 3e^x \, dx - \int_{-1}^{1} x^2 \, dx
\]

1. The integral of \( 3e^x \):
\[
\int_{-1}^{1} 3e^x \, dx = 3 \left[ e^x \right]_{-1}^{1} = 3(e^1 - e^{-1}) = 3(2.718 - 0.3679) = 3(2.3501) = 7.0503
\]

2. The integral of \( x^2 \):
\[
\int_{-1}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}
\]

### Step 5: Calculate the total area
Now, we subtract the results of the two integrals:
\[
\text{Area} = 7.0503 - \frac{2}{3} = 7.0503 - 0.6667 = 6.3836
\]

Thus, the area of the region bounded by the functions \( f(x) = 3e^x - 1 \) and \( g(x) = x^2 - 1 \) on the interval \([-1, 1]\) is approximately \( 6.384 \).

Would you like a detailed explanation on any of the steps? Here are 5 questions to expand on:

1. What happens if the functions intersect more than once on the interval?
2. How do we handle finding intersection points when solving \( 3e^x = x^2 \)?
3. Could we estimate the intersection points numerically?
4. How do we integrate exponential and polynomial functions?
5. What if one of the functions was below the x-axis?

**Tip**: Always check the bounds of integration carefully when working with areas between curves.

Find the area of the region bounded by the functions f(x) = 3e^x - 1 and g(x) = x^2 - 1 over the interval [-1, 1]. This detailed solution walks through setting up the integral, solving for the intersection points, and calculating the area using integration. Ideal for students in Grades 11-12 studying calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation