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Chapter 16: Problem 45

Find the area between the \(y\)-axis and \(y=x^{2}\) from \(y=4\) to \(y=9\). a) \(\frac{29}{3}\) b) \(\frac{32}{3}\) c) \(\frac{34}{3}\) d) \(\frac{38}{3}\)

Short Answer

Expert verified
Question: Find the area between the y-axis and the curve \(y=x^2\) from \(y=4\) to \(y=9\). Answer: \(\frac{19}{3}\)

Step by step solution

01

Define the Function and Bounds

Define the function \(y=x^2\) and the given bounds \(y=4\) and \(y=9\).
02

Find Limits for Integration

We want to find the area between the curve \(y=x^2\) and the \(y\)-axis, so we need to determine the \(x\) values that correspond to \(y=4\) and \(y=9\). Solve the equations \(x^2=4\) and \(x^2=9\) to get these limits: \(x^2=4 \Rightarrow x=\pm 2\) \(x^2=9 \Rightarrow x=\pm 3\) Since we are considering the area between the curve and the \(y\)-axis, we only need the positive values of \(x\). Therefore, we have \(x=2\) and \(x=3\) as the limits for integration.
03

Calculate Area Using Integration

Now, let's calculate the area using integration: Area = \(\int_{2}^{3} x^2 dx\)
04

Integrate the Function

We know that the integral of \(x^2\) is \(\frac{1}{3}x^3\). Let's apply the limits: Area = \(\left[\frac{1}{3}x^3\right]_{2}^{3}\)
05

Evaluate the Area

Now, we evaluate the area using the limits: Area = \(\frac{1}{3}(3^3 - 2^3)\) Area = \(\frac{1}{3}(27-8)\) Area = \(\frac{1}{3}(19)\) Area = \(\frac{19}{3}\) The correct answer is not among the provided options. However, the closest answer is (a) \(\frac{29}{3}\), although it is not the correct result.

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