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Find the area of the region between the curves. $$y=e^{x} \text { and } y=\frac{1}{x^{2}} \text { from } x=1 \text { to } x=2$$

Short Answer

Expert verified
The area is \( e^2 - e - 0.5 \).

Step by step solution

01

Identify the curves

The two curves are given by the functions \( y = e^x \) and \( y = \frac{1}{x^2} \). From the problem, we know the interval is from \( x = 1 \) to \( x = 2 \).
02

Determine the points of intersection

Find where \( y = e^x \) and \( y = \frac{1}{x^2} \) intersect within the given interval. This ensures the correct bounds for integration.
03

Set up the integral

The area between the curves from \( x = 1 \) to \( x = 2 \) is given by the integral of the difference of the functions: \[ A = \int_{1}^{2} \big(e^x - \frac{1}{x^2}\big) \,dx \]
04

Integrate each function separately

Integrate \( e^x \) and \( \frac{1}{x^2} \) separately: \( \int e^x \, dx = e^x \) \( \int -\frac{1}{x^2} \, dx = \int -x^{-2} \, dx = \frac{1}{x} \)
05

Evaluate the definite integrals

Evaluate the integrals from \( x = 1 \) to \( x = 2 \): \( \Big[ e^x \Big]_1^2 = e^2 - e \) \( \Big[ -\frac{1}{x} \Big]_1^2 = -\frac{1}{2} + 1 = 0.5 \)
06

Subtract to find the area

Subtract the integral of \( \frac{1}{x^2} \) from the integral of \( e^x \) to find the area between the curves: \( (e^2 - e) - 0.5 \)
07

Simplify the expression

Simplify the expression to get the final result: \( e^2 - e - 0.5 \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

definite integrals
A definite integral is a fundamental concept in calculus. It allows us to find the exact area under a curve between two points. When calculating definite integrals, we add up tiny rectangles under the curve in a specific interval. The formula for the definite integral of a function \( f(x) \) from \( a \) to \( b \) is written as:
\[ \int_{a}^{b} f(x) \, dx \]
The definite integral has several properties:
  • It gives the net area between the curve and the x-axis.
  • Areas above the x-axis are positive, while those below are negative.
  • We can use it to find areas between two curves by integrating the difference of their functions.

The problem in your exercise uses these properties to find the area between the curves \( y = e^x \) and \( y = \frac{1}{x^2} \) from \( x = 1 \) to \( x = 2 \). To set up the integral, we take the difference of these two functions: \( \int_{1}^{2} \big(e^x - \frac{1}{x^2}\big) \,dx \). This represents the area between these curves in the given interval.
integration
Integration is another core concept in calculus, closely related to differentiation. While differentiation finds the rate of change, integration finds the total accumulation. Integration can be thought of as the reverse process of differentiation. There are different methods of integration, such as:
  • Indefinite Integration: Finding a general form of the antiderivative of a function.
  • Definite Integration: Calculating the exact area under a curve between two points.

For your problem, you used definite integration to find the area between the curves. You first integrated the function \( e^x \) on its own:
\[ \int e^x \, dx = e^x \]
Then, you integrated the function \( -\frac{1}{x^2} \):
\[-\int \frac{1}{x^2} \, dx = \int -x^{-2} \, dx = \frac{1}{x} \]
Evaluating these from \( x = 1 \) to \( x = 2 \) gives you the two areas needed to find the total area between the curves.
exponential functions
An exponential function is a mathematical function of the form \( f(x) = a^x \), where \( a \) is a positive constant. A commonly used base is the natural exponential function \( e^x \), where \( e \) (approximately 2.718) is a unique number called Euler's number. Exponential functions have distinct properties:
  • They grow rapidly as x increases if \( a > 1 \).
  • Their derivative \( f'(x) \) is proportional to the function itself \( f'(x) = a^x \, \ln(a) \).
  • The integral of \( e^x \) is also \( e^x \).

In your exercise, understanding exponential functions helps you integrate \( e^x \) quickly. Specifically, you used:
\[ \int e^x \, dx = e^x \]
This simple property allowed you to focus more on the process of adding and subtracting areas within the definite interval to find the area between the curves.

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Most popular questions from this chapter

We show that, as the number of subintervals increases indefinitely, the Riemann sum approximation of the area under the graph of \(f(x)=x^{2}\) from 0 to 1 approaches the value \(\frac{1}{3},\) which is the exact value of the area. Partition the interval [0,1] into \(n\) equal subintervals of length \(\Delta x=1 / n\) each, and let \(x_{1}, x_{2}, \ldots, x_{n}\) denote the right endpoints of the subintervals. Let $$S_{n}=\left[f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{n}\right)\right] \Delta x$$ denote the Riemann sum that estimates the area under the graph of \(f(x)=x^{2}\) on the interval \(0 \leq x \leq 1\). (a) Show that \(S_{n}=\frac{1}{n^{3}}\left(1^{2}+2^{2}+\cdots+n^{2}\right)\) (b) Using the previous exercise, conclude that $$S_{n}=\frac{n(n+1)(2 n+1)}{6 n^{3}}$$ (c) As \(n\) increases indefinitely, \(S_{n}\) approaches the area under the curve. Show that this area is \(1 / 3 .\)

Use a Riemann sum to approximate the area under the graph of \(f(x)\) on the given interval, with selected points as specified. \(f(x)=x^{3} ; 0 \leq x \leq 1, n=5,\) right endpoints

A property with an appraised value of \(\$ 200,000\) in 2015 is depreciating at the rate \(R(t)=-8 e^{-0.04 t},\) where \(t\) is in years since 2015 and \(R(t)\) is in thousands of dollars per year. Estimate the loss in value of the property between 2015 and 2021 (as \(t\) varies from 0 to 6 ).

Determine \(\Delta x\) and the midpoints of the subintervals formed by partitioning the given interval into \(n\) subintervals. $$1 \leq x \leq 4 ; n=5$$

Use a Riemann sum with \(n=4\) and right endpoints to estimate the area under the graph of \(f(x)=2 x-4\) on the interval \(2 \leq x \leq 3 .\) Then, repeat with \(n=4\) and midpoints. Compare the answers with the exact answer, \(1,\) which can be computed from the formula for the area of a triangle.

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