Problem 9 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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 \(4t^2 - 4t + 1\).

Step 3: 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 \(4/3t^3 - 2t^2 + t\).

Step 4: 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.

Step 5: 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 \(1/3 - 0 = 1/3\).