Question

Dx integral of (t^2 + 1)^10 dt from 0 to x =?

Answer 1

\[
D_x\int_0^xf(t)\;dt
\]First use FTC, where \(F'(t)=f(t)\): \[

D_x\left(F(x)-F(0)\right)=D_xF(x)-D_xF(0)
\]Remember that \(F(0)\) is a constant.\[
D_xF(x)=F'(x)
\]Now if you let \(x=t\) you get\[
F'(x) = f(x)
\]

Answer 2

Thus: \[
D_x\int _a^xf(t)\;dt = f(x)
\]Assuming that \(a\) is some constant with respect to \(x\).

Answer 3

ok..what's FTC? first theorem of calculus?

Answer 4

Fundamental theorem of calculus. Given \(F'(x) = f(x)\) then: \[
\int_a^bf(x)\;dx = F(b)-F(a)
\]

Answer 5

That is how I did: \[
D_x\color{red}{\int_0^xf(t)\;dt} =D_x\color{red}{( F(x)-F(0))}
\]

Answer 6

and so the answer would be (x^2 + 1)^10

Answer 7

Yes.

Answer 8

By the way, \[
D_x\int_a^{g(x)}f(t)\;dt=D_x[F(g(x))-F(a)]=D_xF(g(x)) = F'(g(x))g'(x)
\]

Answer 9

wait how does the Dx F(a) cancel out...

Answer 10

Because \(F(a)\) is a constant with respect to \(x\). Derivatives of constants are \(0\).

Answer 11

i though the integral of f(x) from a to b is F(b)-F(a)

Answer 12

Yeah, in this case \(b=x\) and \(a=0\).

Answer 13

but F(0) is 1...

Answer 14

right? cuz the derivative cancels out the integral
and when u plug in 0 as x u get 1

Answer 15

i mean as t