Question

How do you calculate the value of the integral `inte^(4t²-t) dt` from `[3,x]`?

Answer 1

`inte^(4t^2-t)dt=(e^(4x^2-x))/(8x-1)-e^(33)/23`

Explanation:

Be `f(x)=e^(4t^2-t)` your function.

In order to integrate this function, you will need its primitive `F(x)`

`F(x)=(e^(4t^2-t))/(8t-1)+k` with `k` a constant.

The integration of `e^(4t^2-t)` on [3;x] is calculated as follows:

`inte^(4t^2-t)dt=F(x)-F(3)`

`=(e^(4x^2-x))/(8x-1)+k-((e^(4cdot3^2-3))/(8cdot3-1)+k)`

`=(e^(4x^2-x))/(8x-1)-e^(33)/23`

Answer 2

That integral cannot be expressed using elementary functions. If requires the use of `int e^(x^2) dx`. However the derivative of the integral is `e^(4x^2-x)`

Explanation:

The fundamental theorem pf calculus part 1 tells us that the derivative with respect to `x` of:

`g(x) = int_a^x f(t) dt` is `f(x)`

So the derivative (with respect to `x`) of

`g(x) = int_3^x e^(4t^2-t) dt" "` is `" "g'(x) = e^(4x^2 -x)`.