Question

sketch and find the area of the region determined by the intersection of the curves.
Top - bottom (This is integration-Calculus)

Answer 1

y=x^3 and y=3x+2

Answer 2

my limits are -1 to 2

Answer 3

I did this by hand and got 25/4 but my calculator states it is 27/4

Answer 4

do I have to put this in descending order when I integrate it?
I left it as 3x+2-x^3

Answer 5

a friend of mine used -x^3+3x+2 and got the calculator solution

Answer 6

would something like that throw my solution off?

Answer 7

Well nope..
$$
\int_{-1}^2 (3x + 2 - x^3) dx = \bigg(3\frac{x^2}{2} + 2x - \frac{x^4}{4}\bigg) \bigg|_{-1}^2 = \\
= \bigg(3\frac{2^2}{2} + 2\cdot2 - \frac{2^4}{4}\bigg) - \bigg(3\frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^4}{4}\bigg) = \\
= \bigg(6 + \cancel{4 - 4}\bigg) - \bigg(\frac{3}{2} - 2 - \frac{1}{4}\bigg) = \\
= 6 - \frac{3}{2} + 2 + \frac{1}{4} = 8 - \frac{6}{4} + \frac{1}{4} = \frac{32}{4} - \frac{5}{4} = \frac{27}{4}
$$

Answer 8

I dropped a sign......so typical

Answer 9

thanks so much! I appreciate it

Answer 10

heh, tell me about it =)

Answer 11

sure np