MHB What is the area between the functions $y = |2x|$ and $y = x^2 - 3$?

[tmt1]

Hi,

I need to find the area between these 2 functions:

$$y = |2x|$$

and

$$y = x^2 - 3$$

So I need to find the points of intersection:

$$|2x| - x^2 + 3 = 0$$

for which I get

x = 3, -1

However, since there are no negative x values in y = |2x| I get

$x = 3, 1$

I find that $y = |2x| $is greater than$ y = x^2 - 3$ for this range so

$$\int_{1}^{3} |2x| - x^2 + 3 \,dx$$

So I get

$$\left[x^2 - \frac{x^3}{3} + 3x]\right]_1^3$$

$(9 - 9/3 + 9) - (1 -1/3 + 3)$

and my answer is

$34/3$

However, the answer is 18.

I wasn't sure how to deal with the absolute value exactly so that may be the problem.

 

[Siron]

You've made a mistake in the calculations of the points of interesection. It has to be: $x=3$ and $x=-3$.

To solve problems like this I would recommend to make a plot of the functions to see the area between them more clearly. It'll make it easier for you to spot the integration bounds.

 

[Prove It]

Hi,

I need to find the area between these 2 functions:

$$y = |2x|$$

and

$$y = x^2 - 3$$

So I need to find the points of intersection:

$$|2x| - x^2 + 3 = 0$$

for which I get

x = 3, -1

However, since there are no negative x values in y = |2x| I get

$x = 3, 1$

I find that $y = |2x| $is greater than$ y = x^2 - 3$ for this range so

$$\int_{1}^{3} |2x| - x^2 + 3 \,dx$$

So I get

$$\left[x^2 - \frac{x^3}{3} + 3x]\right]_1^3$$

$(9 - 9/3 + 9) - (1 -1/3 + 3)$

and my answer is

$34/3$

However, the answer is 18.

I wasn't sure how to deal with the absolute value exactly so that may be the problem.

Note that $\displaystyle \begin{align*} \left| 2x \right| = \begin{cases} \phantom{-}2x \textrm{ if } x \geq 0 \\ -2x \textrm{ if } x < 0 \end{cases} \end{align*}$

so you actually have two equations to solve, the first being

$\displaystyle \begin{align*} 2x = x^2 - 3 \end{align*}$ for $\displaystyle \begin{align*} x \geq 0 \end{align*}$ and $\displaystyle \begin{align*} -2x = x^2 - 3 \end{align*}$ for $\displaystyle \begin{align*} x < 0 \end{align*}$.