- #1
tmt1
- 234
- 0
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.
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.