Find values for z for which the function f grows

AndrejN96
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1. Homework Statement


As the title says, I am supposed to find values for x for which the function given below grows.

f(x)=(integral from -3 to x of t^4*e^(t^2)dt)+(integral from x^2 to 2 of t*e^tdt)

Homework Equations

The Attempt at a Solution



I tried solving using substitution or partial integration but I am stuck.
 
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Is this what you have:

##f(x) = \int_{-3}^{x} t^4 e^{t^2} dt + \int_{x^2}^2 te^tdt##

How would you normally work out when a function is increasing?
 
PeroK said:
Is this what you have:

##f(x) = \int_{-3}^{x} t^4 e^{t^2} dt + \int_{x^2}^2 te^tdt##

How would you normally work out when a function is increasing?
I would find the derivative of the function and find for which values of x the value is >0. Totally overlooked it. Thank you.
 
Just to say, i think you might need to put just a tiny bit more caution when calculating the derivative of \int_{x^2}^{2}te^tdt wrt x.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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