Problem 62
Question
Find \(F\) as a function of \(x\) and evaluate it at \(x=2, x=5,\) and \(x=8\). $$ F(x)=\int_{2}^{x}\left(t^{3}+2 t-2\right) d t $$
Step-by-Step Solution
Verified Answer
The evaluated function \(F(x)\) at \(x = 2, x = 5, x = 8\) is 0, 120.75, and 1016 respectively.
1Step 1: Calculating the Integral
First, calculate the given integral. Using the power rule for integration \(\int x^n dx = 1/(n+1) * x^{n+1}\) and the fact that the integral of a constant is just the constant times the variable, the integral of \(F(x)=\int_{2}^{x}\left(t^{3}+2 t-2\right) d t\) results in \(F(x) = 1/4*(x^4 - 2^4) + x^2 -2*(x-2)\).
2Step 2: Evaluating F(x) at x = 2
Substitute \(x = 2\) into the function. We obtain \(F(2) = 1/4*(2^4 - 2^4) + 2^2 -2*(2-2) = 0\).
3Step 3: Evaluating F(x) at x = 5
Substitute \(x = 5\) into the function. We obtain \(F(5) = 1/4*(5^4 - 2^4) + 5^2 -2*(5-2) = 120.75\).
4Step 4: Evaluating F(x) at x = 8
Substitute \(x = 8\) into the function. We obtain \(F(8) = 1/4*(8^4 - 2^4) + 8^2 -2*(8-2) = 1016\).
Key Concepts
Integration TechniquesEvaluating IntegralsPower Rule for Integration
Integration Techniques
Understanding integration techniques is crucial for solving calculus problems involving integrals. These techniques are methods or tools that enable students to tackle a wide variety of functions that they need to integrate. When you approach an integral, such as
\[ F(x)=\int_{2}^{x}\left(t^{3}+2 t-2\right) d t \],
you first look for a method that applies to the integrand (the function you are integrating). In our example, we're dealing with a polynomial, which is one of the simpler kinds of functions to integrate.
\[ F(x)=\int_{2}^{x}\left(t^{3}+2 t-2\right) d t \],
you first look for a method that applies to the integrand (the function you are integrating). In our example, we're dealing with a polynomial, which is one of the simpler kinds of functions to integrate.
Common Integration Techniques
- Power Rule: Useful for integrating polynomial terms like \( t^n \).
- Substitution: Helpful when dealing with compositions of functions or when a change of variables simplifies the integral.
- Integration by Parts: Applies to products of functions, based on the product rule for differentiation.
Evaluating Integrals
The process of evaluating integrals involves finding the antiderivative of the given function and then calculating the values over the specified interval. The definite integral, which has both upper and lower limits, gives you the net area under the curve of the function between those limits.
When you have the integral
\[ F(x)=\int_{2}^{x}\left(t^{3}+2 t-2\right) d t \],
you are asked to find the net area from \( t = 2 \) to \( t = x \). After you find the antiderivative, which is, in essence, the function \( F(x) \), you can evaluate it at various points, such as \( x = 2 \), \( x = 5 \), and \( x = 8 \), to get specific values.
Evaluating at a point involves substituting the values into the antiderivative. For instance, when we plug in \( x = 2 \), the integral evaluates to 0 because it represents the area from a point to itself, which is zero. This step is crucial to understanding how the function behaves over different intervals and helps visualize the grown or shrinkage across the range of the integral.
When you have the integral
\[ F(x)=\int_{2}^{x}\left(t^{3}+2 t-2\right) d t \],
you are asked to find the net area from \( t = 2 \) to \( t = x \). After you find the antiderivative, which is, in essence, the function \( F(x) \), you can evaluate it at various points, such as \( x = 2 \), \( x = 5 \), and \( x = 8 \), to get specific values.
Evaluating at a point involves substituting the values into the antiderivative. For instance, when we plug in \( x = 2 \), the integral evaluates to 0 because it represents the area from a point to itself, which is zero. This step is crucial to understanding how the function behaves over different intervals and helps visualize the grown or shrinkage across the range of the integral.
Power Rule for Integration
The power rule for integration is one of the most fundamental tools in calculus. It applies to functions of the form \( t^n \) and states that the integral of \( t^n \) with respect to \( t \) is \( \frac{1}{n+1}t^{n+1} \), provided that \( n \) is not equal to -1. In our problem, the integral of \( t^3 \) would be \( \frac{1}{4}t^4 \) following the power rule. But it's equally important to remember to subtract the lower limit from the upper limit's result after applying the power rule in a definite integral.
Take a look at the exercise, where this method simplifies the process considerably:
Take a look at the exercise, where this method simplifies the process considerably:
- For the term \( t^3 \), apply the power rule to get \( \frac{1}{4}t^4 \).
- For the linear term \( 2t \), recall that the antiderivative is \( t^2 \).
- And for the constant \( -2 \), its antiderivative is \( -2t \).
Other exercises in this chapter
Problem 62
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True or False? In Exercises 61 and 62, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.
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