Find the area bounded by the curves $y = e^x$ and $y = x^2 - 2$ from x = -1 to x = 1

Question

anonymous · Answer

$$\int\limits_{-1}^{1}e^x - (x^2-4)dx = (e^x -\frac{x^3}{3} +4x)|_{-1}^{1}$$
=$$(e - \frac{1}{3}+4)-(\frac{1}{e}+\frac{1}{3}-4) = 8-\frac{1}{e}+e$$

lgbasallote · Answer

$$\large e - \frac{1}{3} + 4 - \frac{1}{e} - \frac{1}{3} + 4$$ then it should be $$\large e - \frac{1}{2} - \frac{2}{3} + 8$$ @SmoothMath

lgbasallote · Answer

although i think it should be 2 not 4

lgbasallote · Answer

i dont recall the question well lol

lgbasallote · Answer

uhh that was supposed to be $$\large e - \frac{1}{e} - \frac{2}{3} + 8$$

anonymous · Answer

Ah, yes. Correct. And why would it be 2? 4x is the anti-derivative of 4.

lgbasallote · Answer

no...i meant the question should be the are bounded by $y = e^x$ and $y  =x^2 -2$ not 4...anyway...ill just see the process here...what comes next?

anonymous · Answer

I mean... nothing really. You can rewrite that if you like though.

lgbasallote · Answer

so here was my solution.... $$\int_{-1}^1 e^x -(x^2 - 2)dx$$ $$\int_{-1}^1 e^x - x^2  + 2$$ $$[e^x|_{-1}^1 - [\frac{x^3}{3}|_{-1}^1 + [2x|_{-1}^1$$ $$[e - \frac{1}{e}] - [\frac{1}{3} - (-\frac{1}{3})] + [2 - (-2)]$$ $$(\frac{e^2 - 1}{e}) - \frac{2}{3} + 4$$ $$\frac{3(e^2 -1) - 2e + 12e}{3e} = \frac{3(e^2 - 1) + 10e}{3e}$$ 

that right?? @smoothmath ?

anonymous · Answer

Quite right. =) Yeah, I changed it to 4 for some reason. Silly me.

lgbasallote · Answer

no it was originally 4 lol =))

lgbasallote · Answer

it seems that was the question

anonymous · Answer

or you could simplify that down to
$$\frac{3e^2 + 10e - 3}{3e}$$

anonymous · Answer

Oh, hahaha

lgbasallote · Answer

thanks...that was a great reassurance that i can actually pass the subject :DDDD