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Chapter 4: Problem 9

Evaluate the definite integral of the algebraic function. Use a graphing utility to verify your result. $$ \int_{0}^{1}(2 t-1)^{2} d t $$

Short Answer

Expert verified
The value of the definite integral of the given function from 0 to 1 is 1/3.

Step by step solution

01

Identify the function and limits of integration

The function given to us for integration is \((2t - 1)^2\) and the limits of integration are 0 and 1. This means we want to evaluate the area under the curve from t=0 to t=1.
02

Rewrite the function

Let's rewrite the function in a way that makes integration simpler. Instead of \((2t - 1)^2\), let's rewrite the function as \(4t^2 - 4t + 1\).
03

Integrate

Now, we integrate term by term. The antiderivative of \(4t^2\) is \(4/3t^3\), the antiderivative of \(-4t\) is \(-2t^2\), and the antiderivative of \(1\) is \(t\). So, the antiderivative of \(4t^2 - 4t + 1\) is \(4/3t^3 - 2t^2 + t\).
04

Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus tells us that to find the definite integral over the interval [0, 1], we subtract the value of the antiderivative at the lower limit from the value at the upper limit. Let's do this calculation.
05

Substitute the limits of integration

Substituting \(t = 1\) into \(4/3t^3 - 2t^2 + t\), we get \(4/3 - 2 + 1 = 1/3\). Substituting \(t = 0\), we get \(0\). Therefore, the value of the definite integral over the interval [0, 1] is \(1/3 - 0 = 1/3\).

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Most popular questions from this chapter

Evaluate the integral. \(\int_{0}^{4} \frac{1}{\sqrt{25-x^{2}}} d x\)

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