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Chapter 3: Problem 2

Find area between curves \(\mathrm{y}=\mathrm{x}^{2}\) and \(\mathrm{y}=3 \mathrm{x}-2\) from \(\mathrm{x}=0\) to \(\mathrm{x}=2\).

Short Answer

Expert verified
Answer: The area between the curves is 4/3 square units.

Step by step solution

01

Determine the intersection points of the curves

First, we need to find the intersection points of the two curves. We can do this by setting \(\mathrm{y}=\mathrm{x}^{2}\) equal to \(\mathrm{y}=3\mathrm{x}-2\). \(\mathrm{x}^2 = 3\mathrm{x}-2\) After solving this equation, we can determine the intersection points of the curves.
02

Solve the equation for intersection points

To solve the equation from step 1, we can move all terms to one side: \(\mathrm{x}^2 - 3\mathrm{x}+2=0\) This is a quadratic equation that can be factored as \((\mathrm{x}-1)(\mathrm{x}-2)=0\). The solutions are \(\mathrm{x}=1\) and \(\mathrm{x}=2\). So, the area of interest is bound by the points \(\mathrm{x}=0\), \(\mathrm{x}=1\), and \(\mathrm{x}=2\).
03

Set up the definite integral for the area between the curves

To find the area between the curves, we can set up a definite integral by subtracting the lower curve function from the higher curve function and integrating from the leftmost bound to the rightmost bound: \(\int_{a}^{b} (y_\text{upper} - y_\text{lower})\, \mathrm{d}\mathrm{x}\) Since the curve \(\mathrm{y}=3\mathrm{x}-2\) is above \(\mathrm{y}=\mathrm{x}^{2}\) for \(0\le x \le 2\), the integral becomes: \(\int_{0}^{2} (3\mathrm{x}-2-\mathrm{x}^2) \, \mathrm{d}\mathrm{x}\)
04

Evaluate the definite integral

To evaluate the definite integral, we integrate the function with respect to x: \(\int(3\mathrm{x}-2-\mathrm{x}^2) \, \mathrm{d}\mathrm{x} = \left[\frac{3}{2}\mathrm{x}^2-2\mathrm{x}-\frac{1}{3}\mathrm{x}^3\right]\) Now, we apply the limits of integration by evaluating the antiderivative at the upper bound, 2, and subtracting the evaluation at the lower bound, 0: \(\left[\frac{3}{2}(2)^2-2(2)-\frac{1}{3}(2)^3\right] - \left[\frac{3}{2}(0)^2-2(0)-\frac{1}{3}(0)^3\right]=4-4-\frac{8}{3}=-\frac{4}{3}\) Due to the negative value, the final result will be the absolute value of the calculated area.
05

Take absolute value

Since the area cannot be negative, we take the absolute value: \(|\text{-}\frac{4}{3}| = \frac{4}{3}\) Thus, the area between the curves \(\mathrm{y}=\mathrm{x}^{2}\) and \(\mathrm{y}=3\mathrm{x}-2\) from \(\mathrm{x}=0\) to \(\mathrm{x}=2\) is \(\frac{4}{3}\).

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