Problem 81 - Calculus Early Transcendentals

Question

Find the area of the following regions. The region bounded by the graph of \(f(x)=(x-4)^{4}\) and the \(x\) -axis between \(x=2\) and \(x=6\)

Step-by-Step Solution

Answer

Answer: The area of the region is \(\frac{64}{5}\).

Step 1: Find the integral of f(x)

First, we need to find the integral of the function \(f(x)=(x-4)^{4}\). Since it's a polynomial, we can simply apply the power rule for integration: \(\int (x-4)^{4} dx = \frac{(x-4)^{5}}{5} + C\)

Step 2: Set up the definite integral

Now, we need to set up the definite integral that corresponds to the area bounded by the graph of the function, the \(x\)-axis, and the vertical lines \(x=2\) and \(x=6\). This can be written as: \(A = \int_{2}^{6} (x-4)^{4} dx \)

Step 3: Evaluate the definite integral

Next, we'll evaluate this integral by applying the fundamental theorem of calculus. That is, we need to find the antiderivative of our function, which is the function we found in step 1, and evaluate it at the upper and lower limits, then subtract the lower from the upper: \(A = \left[\frac{(x-4)^{5}}{5}\right]_{2}^{6}\)

Step 4: Substitute the bounds

We substitute the bounds (2 and 6) into our antiderivative and then subtract them: \(A = \frac{(6-4)^{5}}{5} - \frac{(2-4)^{5}}{5}\) Evaluate the expression in the parentheses: \(A = \frac{2^{5}}{5} - \frac{(-2)^{5}}{5}\)

Step 5: Simplify the expression

Finally, simplify the expression to find the area: \(A = \frac{32}{5} - \frac{-32}{5}\) \(A = \frac{32+32}{5}\) \(A = \frac{64}{5}\) Hence, the area of the region bounded by the graph of the function, the x-axis, and the vertical lines \(x=2\) and \(x=6\) is \(A=\frac{64}{5}\).