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Chapter 5: Problem 11

Find the area under the given curve over the indicated interval. $$ y=e^{x} ; \quad[0,3] $$

Short Answer

Expert verified
The area under the curve \(y = e^x\) from 0 to 3 is \(e^3 - 1\).

Step by step solution

01

- Identify the Integral

To find the area under the curve, integrate the function over the given interval. For the function \(y = e^x\) over the interval \([0,3]\), set up the integral as: \[ \text{Area} = \int_{0}^{3} e^x \, dx \]
02

- Perform the Integration

Find the antiderivative of the integrand. The antiderivative of \(e^x\) is simply \(e^x\). This gives: \[ \int e^x \, dx = e^x + C \]
03

- Apply the Limits of Integration

Evaluate the antiderivative from 0 to 3. Use the Fundamental Theorem of Calculus: \[ \text{Area} = \left[e^x\right]_0^3 = e^3 - e^0 \]
04

- Simplify the Expression

Simplify the expression to find the exact area. Since \(e^0 = 1\), the area under the curve is: \[ e^3 - 1 \]

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Key Concepts

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

integration
Integration is a core concept in calculus. It helps find the area under a curve. When we integrate a function, we're essentially summing up an infinite number of infinitesimally small areas (like tiny rectangles) under the curve. In our example, we need to find the area under the curve defined by the exponential function\(y = e^x\). The integral we'll compute is: \[ \text{Area} = \int_{0}^{3} e^x \, dx \]
This tells us to add up all values of the exponential function from 0 to 3.
antiderivative
An antiderivative is the reverse of a derivative. It is a function whose derivative gives back the original function. Finding the antiderivative is also known as 'indefinite integration'. For the function\(e^x\), its antiderivative is also \(e^x\).
Mathematically, we can express this as: \[\int e^x \, dx = e^x + C \] Here, \(C\) is an arbitrary constant which we get when we integrate any function indefinitely. When we apply limits (as we'll do in our example), the \(C\) cancels out, so we don't need to worry about it for definite integration.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects differentiation and integration. It states two important things:
1. If you take the integral of a function, then differentiate it, you get back the original function.
2. If you have an integral with limits of integration, you can evaluate it using the antiderivative.
Formally, if\(F(x)\) is an antiderivative of\(f(x)\), then: \[\int_{a}^{b} f(x) \, dx = F(b) - F(a) \]
In our example: 鈭玣rom 0 to 3 of \(e^x \ dx\) = \[ e^3 - e^0\]. Note that \(e^0\) simplifies to 1.
exponential function
The exponential function\(e^x\) is one of the most important functions in mathematics. It's defined as the function where the rate of change at any point is proportional to the value of the function at that point. This is why its derivative and antiderivative are also\(e^x\).
Some important properties include:
  • \(e^0 = 1\)
  • \(e^x \, e^y = e^{x + y}\)
  • \( (e^x)^n = e^{nx}\)
This makes\(e^x\) very predictable and easy to work with, which is why it's so commonly used in calculus.

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Most popular questions from this chapter

The rate of electrical energy used by a family, in kilowatts, is given by $$ K(t)=10 t e^{-t} $$ where \(t\) is the time, in hours. That is, \(t\) is in the interval \([0,24]\). a) How many kilowatt hours does the family use in the first \(\mathrm{T}\) hours of a day \((t=0\) to \(t=T) ?\) b) How many kilowalt hours does the family use in the first \(4 \mathrm{hr}\) of the day?

Evaluate using integration by parts. $$ \int_{0}^{5 \pi / 6} 3 x \cos x d x $$

a) Integrate \(\int x \sqrt[3]{x+2} d x\) using a substitution. b) Use integration by parts to compute \(\int x \sqrt[3]{x+2} d x\) c) Use algebra to show that your answers in parts (a) and (b) are equal.

The annual divorce rate in the United States is approximated by $$ D(t)=100,000 e^{0.025 t}, $$ where \(D(t)\) is the divorce rate at time \(t\) and \(t\) is the number of years measured from 1900 . That is, \(\ell=0\) corresponds to \(1900, t=98 \frac{9}{365}\) corresponds to January 9,1998, and so on. a) Find the total number of divorces from 1900 to 1999\. Note that this is $$ \int_{0}^{99} D(t) d t. $$ b) Find the total number of divorces from 1980 to \(1999 .\) Note that this is $$ \int_{80}^{99} D(t) d t. $$

Evaluate using integration by parts. Check by differentiating. $$ \int x e^{-2 x} d x $$
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