Kindergarten Homework: 10S t^3 dt=f(x)

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HI people, this will be kindergarden homework for you guys. PLease help! THANKS ALOT!

S=integral sign

10
S t^3 dt=f(x)
x

f'(x)=?
I say f'(x)=t^3, but it's not correct. WHY?
 
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First, find the indefinite integral of t3 dt, then evaluate it from t=x to t=10. Only after you've completed that can you differentiate the resultant expression with respect to x.

- Warren
 
so the indefinite integral of t^3 dt is t^4/4

then evaluate it from t=x to t=10

10^4/4-x^4/4

then differentiate this expression

0-x^3

it's correct!... I am dumb THANX ALOT fOR The fAsT repLY !
 
Last edited:
You're not dumb at all. See, you just learned something. :)

- Warren
 
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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