Question

Let f(x) = integral e^(t^2) dt from -2 to x^2 - 3x. At what value of x is f(x) a minimum?

Answer 1

extremum occurs when f'(x)=0

Answer 2

But you should evaluate the integral first

Answer 3

I got lost with all the exponents lol.

Answer 4

Try integral by part

Answer 5

u have to use leibniz's rule

Answer 6

Well I can do it without the leib- whatever rule

Answer 7

how did the variables disappear :O

Answer 8

it'll be easy with it

Answer 9

1/2sqrt(pi)e(t) ? Lol, idk ,_,

Answer 10

From where did the pi emerge :O

Answer 11

take a look
http://en.wikipedia.org/wiki/Leibniz_integral_rule

Answer 12

\[\dfrac{d}{dx}\int \limits_{-2}^{x^2-3x }e^{(t^2)}dt = 0\]

Answer 13

use FTC+chain rule

Answer 14

\[\implies \dfrac{d}{dx}\left[f(x^2-3x) - f(-2)\right]= 0\]

\[\implies f'(x^2-3x)(x^2-3x)' -0= 0\]

\[\implies e^{(x^2-3x)^2}(2x-3) = 0\]

\[\implies 2x-3 = 0\]

Answer 15

3/2