Question

Find dy/dx if y=int(t^2+5t-3t)dt interval [2x, 0]

Answer 1

\[\int\limits_{0}^{2x}t^{2}+5t-3\]

Answer 2

Okay, so let's deal with the integral first

Answer 3

that would be t^3/3+5t^2/2-3t

Answer 4

Don't forget the limits!

Answer 5

what to do with them?

Answer 6

\[\Large\left[\frac{t^3}3+5\frac{t^2}2-3t\right]_0^{2x}\]

Answer 7

Now let's do the substitution

Answer 8

[(2x)^3/3+5(2x)^2/2-3(2x)]-0

Answer 9

Expand it

Answer 10

how lol

Answer 11

(2x)^3 = 8x^3

Answer 12

oh like simplify?

Answer 13

Yep sorry

Answer 14

What? Let's not do that. Generally speaking, we have:

If \(\dfrac{d}{dt}F(t) = f(t)\)

Then, \(\int\limits_{a}^{b}f(t)\;dt = F(b) - F(a)\)

Then, \(\int\limits_{0}^{2x}f(t)\;dt = F(2x) - F(0)\)

Then, \(\dfrac{d}{dx}\int\limits_{0}^{2x}f(t)\;dt = \dfrac{d}{dx}\left(F(2x) - F(0)\right) = f(2x)\cdot 2 = 2\left((2x)^{2}+5(2x)−3\right)\)

The significance of this is that we need only THEORIZE that \(F(t)\) exists. We don't actually have to find \(F(t)\). In addition, there's a lot less to go wrong.