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Chapter 11: Problem 28

Evaluate the definite integral. $$\int_{1}^{2}\left(t^{5}-t^{3}+1\right) d t$$

Short Answer

Expert verified
The value of the definite integral \(\int_{1}^{2}\left(t^{5}-t^{3}+1\right) dt\) is \(\frac{125}{12}\).

Step by step solution

01

Find the antiderivative

To evaluate the definite integral, we first need to find the antiderivative of the given function, \(f(t) = t^5 - t^3 + 1\). We'll use the power rule to obtain the antiderivative: Power rule: \(\int t^n dt = \frac{t^{n+1}}{n+1} + C\) Using the power rule for each term, we get: \(\int (t^5 - t^3 + 1)dt = \frac{t^6}{6} - \frac{t^4}{4} + t + C\)
02

Apply the Fundamental Theorem of Calculus

Now, we'll apply the Fundamental Theorem of Calculus to find the definite integral between the bounds 1 and 2: \(\int_{1}^{2} (t^{5} - t^{3} + 1) dt = \left(\frac{t^6}{6} - \frac{t^4}{4} + t\right)\Big|_1^2\)
03

Evaluate using the bounds

Plug in the upper bound, 2, and then subtract the result of plugging in the lower bound, 1: \(\left(\frac{(2)^6}{6} - \frac{(2)^4}{4} + 2\right) - \left(\frac{(1)^6}{6} - \frac{(1)^4}{4} + 1\right)\) Simplify the expression: \(= \left(\frac{64}{6} - 4 + 2\right) - \left(\frac{1}{6} - \frac{1}{4} + 1\right)\) Calculate the final result: \(= \left(\frac{128}{6}\right) - \left(\frac{1}{4}\right)\) \(= \frac{128 - 3}{12}\) \(= \frac{125}{12}\)
04

Final answer

The value of the definite integral \(\int_{1}^{2} (t^{5} - t^{3} + 1) dt\) is \(\frac{125}{12}\).

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Key Concepts

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

Antiderivative
An antiderivative is a function whose derivative is the original function we started with. Think of it as reverse engineering the process of differentiation.When you find the antiderivative of a function, you are discovering a formula that, when differentiated, returns the original function. For example, if we start with the function \( f(t) = t^5 - t^3 + 1 \), we need to find an expression \( F(t) \) such that \( F'(t) = f(t) \). This expression is the antiderivative.
  • The antiderivative of \( t^5 \) is \( \frac{t^6}{6} \). We increase the power by one and divide by the new power.
  • For \( t^3 \), the antiderivative is \( \frac{t^4}{4} \).
  • The constant \( 1 \) becomes \( t \) since the derivative of \( t \) is \( 1 \).
By calculating these, we obtain the full antiderivative: \( \frac{t^6}{6} - \frac{t^4}{4} + t + C \), where \( C \) represents the constant of integration.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a powerful tool that connects differentiation with integration. It has two main parts.First, it tells us that if we have an antiderivative of a function \( f \), we can evaluate the definite integral of \( f \) over an interval \([a, b]\) by calculating the difference \( F(b) - F(a) \), where \( F \) is the antiderivative of \( f \). This allows us to calculate the area under the curve defined by \( f(t) \).In practice, this means:
  • Find the antiderivative \( F(t) \) of the function you are integrating.
  • Evaluate \( F \) at the upper bound of the integral to get \( F(b) \).
  • Evaluate \( F \) at the lower bound to obtain \( F(a) \).
  • Subtract \( F(a) \) from \( F(b) \) to find the value of the definite integral.
In the original exercise, we applied this theorem to calculate:\[ \int_{1}^{2} (t^{5} - t^{3} + 1) dt = \left(\frac{t^6}{6} - \frac{t^4}{4} + t\right)\Big|_1^2 \] with the bounds explicit in the calculation.
Power Rule
The power rule is a fundamental technique used in calculus for finding the antiderivative (or derivative, in reverse). It's very straightforward.The rule is: for any term \( t^n \), where \( n \) is a real number, the antiderivative is \( \frac{t^{n+1}}{n+1} + C \), assuming \( n eq -1 \). This rule helps break down complex polynomials into simple computations. For example, in the exercise, we used it to find:
  • \( \frac{t^6}{6} \) as the antiderivative of \( t^5 \)
  • \( \frac{t^4}{4} \) as the antiderivative of \( t^3 \)
  • \( t \) as the antiderivative of the constant \( 1 \)
Working term by term and adding constants allows you to solve integrals with confidence. Remember this rule as it simplifies many integral problems involving polynomial expressions.

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Most popular questions from this chapter

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(f\) and \(g\) are continuous on \([a, b]\) and either \(f(x) \geq g(x)\) for all \(x\) in \([a, b]\) or \(f(x) \leq g(x)\) for all \(x\) in \([a, b]\), then the area of the region bounded by the graphs of \(f\) and \(g\) and the vertical lines \(x=a\) and \(x=b\) is given by \(\int_{a}^{b}\lfloor f(x)-g(x) \mid d x\).

Sketch the graph and find the area of the region bounded by the graph of the function \(f\) and the lines \(y=0, x=a\), and \(x=b\) $$f(x)=-x^{2}+4 x-3 ; a=-1, b=2$$

Evaluate \(\int_{3}^{0} f(x) d x\), given that \(\int_{0}^{3} f(x) d x=4\).

Sketch the graph and find the area of the region completely enclosed by the graphs of the given functions \(f\) and \(g\). $$f(x)=2 x\( and \)g(x)=x \sqrt{x+1}$$

The total vacancy rate (in percent) of offices in Manhattan from 2000 through 2006 is approximated by the function $$ \begin{array}{r} R(t)=0.032 t^{4}-0.26 t^{3}-0.478 t^{2}+5.82 t+3.8 \\ (0 \leq t \leq 6) \end{array} $$ where \(t\) is measured in years, with \(t=0\) corresponding to 2000 . What was the average vacancy rate of offices in Manhattan over the period from 2000 through \(2006 ?\)
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