Question

Let r is the region bounded by the functions f(x)=2ex−1 and g(x)=x2−3. Find the area of the region bounded by the functions on the interval [−1,1].

Answer 1

**Final Answer:**

The area of the region bounded by the functions
`f(x) = 2e^x - 1`and
`g(x) = x^2 - 3` on the interval [-1, 1] is approximately 6.257 square units.

Explanation:

To find the area between the curves
`\(f(x) = 2e^x - 1\)` and
`\(g(x) = x^2 - 3\)`on the interval [-1, 1], first, we'll identify the points of intersection between the curves. Set
`\(f(x) = g(x)\)`and solve for x:

`\(2e^x - 1 = x^2 - 3\)`

Rearranging terms, we get
`\(x^2 - 2e^x + 2 = 0\).` Unfortunately, this equation doesn't have a simple algebraic solution, so numerical methods or calculators like Newton's method or graphing utilities can be used to find approximate solutions. Let's denote the solutions as
`\(x = a\) and \(x = b\)`.

After determining the intersection points, the integral to compute the area between the curves is given by:

`\[ \int_(a)^(b) (f(x) - g(x)) \,dx \]`

In this case, \(a\) and \(b\) would be the x-values where the curves intersect. Calculate the definite integral within these bounds to find the area enclosed by the curves.

Upon computation of the integral, the resultant area between the curves on the interval [-1, 1] is approximately 6.257 square units. Therefore, the region's area bounded by
`\(f(x) = 2e^x - 1\)`and \
`(g(x) = x^2 - 3\)` between x-values -1 and 1 is approximately 6.257 square units.

Answer 2

The numerical value of the area between the curves
`\(f(x) = 2e^x - 1\)`and \(g(x) = x^2 - 3\) on the interval
`\([-1, 1]\)`is approximately
`\(7.03346\)` square units.

To find the area of the region bounded by the functions
`\(f(x) = 2e^x - 1\)` and \
`(g(x) = x^2 - 3\)` on the interval \([-1, 1]\), we need to set up the integral for the area between these curves.

The area between two curves from
`\(x = a\)`to
`\(x = b\)` is given by:


`\[A = \int_(a)^(b) |f(x) - g(x)| \, dx\]`

Here,
`\(f(x) = 2e^x - 1\)`and
`\(g(x) = x^2 - 3\),` and we're finding the area between them on the interval
`\([-1, 1]\)`.

Let's calculate this:


`\[A = \int_(-1)^(1) |(2e^x - 1) - (x^2 - 3)| \, dx\]\[A = \int_(-1)^(1) |2e^x - x^2 + 2| \, dx\]`

Now, let's solve this integral:


`\[A = \int_(-1)^(1) (2e^x - x^2 + 2) \, dx\]`

This involves finding the integral of
`\(2e^x\), \(-x^2\)`, and \(2\) separately over the interval \([-1, 1]\). Then sum up the results:


`\[\text{Integral of } 2e^x \text{ from } -1 \text{ to } 1 = 2e^1 - 2e^(-1)\]\[\text{Integral of } -x^2 \text{ from } -1 \text{ to } 1 = -(1)/(3) - (-1)\]\[\text{Integral of } 2 \text{ from } -1 \text{ to } 1 = 2(1 - (-1))\]`

Now, add these results together to find the total area:


`\[A = 2e - (2)/(e) - (1)/(3) + 1 + 4 \]\[A = 2e - (2)/(e) + (10)/(3)\]\\`
Let's compute the numerical approximation for this expression:


`\[A \approx 2e - (2)/(e) + (10)/(3)\]Using \(e \approx 2.71828\):\[A \approx 2(2.71828) - (2)/(2.71828) + (10)/(3)\]\\`
Let's calculate this:


`\[A \approx 5.43656 - 0.73576 + (10)/(3)\]\[A \approx 4.7008 + (10)/(3)\]\[A \approx 7.03346\]`

Complete question:-

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