Recall from the First Fundamental Theorem, that if `F(x) = int_a^xf(t)dt`, then `F'(x)=f(x)`.

In this example, `f(t) = t^2+3t-4`.

So:

`d/dx int_5^x (t^2 + 3t - 4)dt = x^2 + 3x - 4`.

That's all there is too it. We don't need to integrate the expression after the integral sign (the integrand) first, then differentiate the result. We can write down the derivative immediately.

However, let's do it the long way round to see how it works.

`int_5^x (t^2 + 3t - 4)dt = [t^3/3 + (3t^2)/2 - 4t]_5^x`

`=[x^3/3 + (3x^2)/2 - 4x ] -` ` [5^3/3 + (3(5)^2)/2 - 4(5)]`

`=[x^3/3 + (3x^2)/2 - 4x] - [59.167]`

`=x^3/3 + (3x^2)/2 - 4x - 59.167`

Next, we take the derivative of this result, with respect to `x`:

`d/dx(x^3/3 + (3x^2)/2 - 4x - 59.167) ` `= x^2 +3x - 4`

This is the same result we obtained before. Notice it doesn't matter what the lower limit of the integral is (in this case, `5`), since the constant value it produces (in this case, `59.167`) will disappear during the differentiation step.

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