Finding the area of a curve where one of the curves is a line

Question

The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney.
Problem:
Find the area of the region bounded by the given curves. $$ y^2 = 4x, y = 4x - 2 $$
Answer:
\begin{align*} 4x &= (4x - 2)^2 \\ 4x &= 16x^2 - 16x + 4 \\ 16x^2 - 20x + 4 &= 0 \\ 4x^2 - 5x + 1 &= 0 \\ x &= \frac{5 \pm \sqrt{25- 4(4)(1)} }{2(4)} = \frac{5 \pm \sqrt{25- 16} }{8} \\ x &= \frac{ 5 \pm 3 }{8} \\ x = 1 &\text{ or } x= \frac{1}{4} \\ \end{align*} Let $A$ be the area we seek. \begin{align*} A &= \int_{ \frac{1}{4} }^{1} 2x ^ \frac{1}{2} - (4x - 2 ) \, dx = \int_{ \frac{1}{4} }^{1} 2x ^ \frac{1}{2} - 4x + 2 \, dx \\ A &= 2 \left( \frac{2}{3} \right) x ^{\frac{3}{2}} - 2x^2 + 2x \Big|_{\frac{1}{4}}^{1} \\ A &= \frac{4}{3} x ^{\frac{3}{2}} - 2x^2 + 2x \Big|_{\frac{1}{4}}^{1} \\ A &= \left( \frac{4}{3}\right) \left( 1 \right) - 2(1) + 2(1) - \left( \left( \frac{4}{3}\right) \left( \frac{1}{4} \right)^{\frac{3}{2}} - \frac{2}{16} + \frac{2}{4 }\right) \\ A &= \frac{4}{3} - \left( \left( \frac{4}{3}\right) \left( \frac{1}{2} \right)^{3} - \frac{1}{8} + \frac{1}{2 }\right) \\ A &= \frac{4}{3} - \left( \frac{4}{3}\right) \left( \frac{1}{8} \right) + \frac{1}{8} - \frac{1}{2} \\ A &= \frac{4}{3} - \frac{4}{24} - \frac{3}{8} = \frac{4(8) - 4 - 3(3)}{24} \\ A &= \frac{19}{24} \\ \end{align*} The book's answer to this problem is $\frac{9}{8}$. I am wondering what I did wrong. I checked my integration with an online integral calculator.

Answer 1

If you actually look at a graph of the required area you will see that it is quite difficult to find when integrating with respect to $x$. We can instead rearrange both equations to get $$x=\frac14y^2$$ $$x=\frac14y+\frac12$$ Then the graphs intersect at points where $y=-1$ and $y=2$ respectively so the area is given by $$\int_{-1}^2\left(\frac14y+\frac12\right)-\left(\frac14y^2\right)\mathrm{d}y=\frac98$$

Answer 2

Your error lies in the computation of the intersection points of the curves. If $y^2=4x$ and $y=4x-2$, then $4x=(4x-2)^2$ indeed, but this is not an equivalence. It turns out that your region is the region below the graph of $2\sqrt x$ and above the graph of $-2\sqrt x$ (with $x\in\left[0,\frac14\right]$) plus the area of the region below the graph of $2\sqrt x$ and above the line $y=4x-2$ (with $x\in\left[\frac14,1\right]$); see the picture below. And the answer is indeed $\frac98$.