1. ## Antiderivative

Which is the easiest method to find the antiderivative of (2t-3).

I know that this may be an easy one, I just need to see ways of finding out how to find an antiderivative.

2. Originally Posted by katchat64
Which is the easiest method to find the antiderivative of (2t-3).

I know that this may be an easy one, I just need to see ways of finding out how to find an antiderivative.
Integrate it with respect to t!

$\int 2t-3 dt = \frac{2t^2}{2} - 3t + C$

$= t^2 - 3t + C$

3. Ok.... I already have to do that, so then it just makes the problem a lot easier right?

4. Say I need to substitute 2 & 0, how would the problem look?

5. If you're a beginner, it helps to write every term (if possible) in the form

$ax^n$.

Then you can remember that the antiderivative is

$\frac{a}{n + 1}x^{n + 1} + C$.

So in this case, you'd have $2t^1 + 3t^0$.

Can you find this antiderivative using the formula?

If you're finding a definite integral, once you have the antiderivative, substitute the upper limit and see what you have ( $F(b)$, substitute the lower limit ( $F(a)$) and see what you have, then do $F(b) - F(a)$.

6. ## Evaluate the Integral

$\int_0^2{(2t-3)\,dx}$

What I got for the antiderivative of (2t-3) is (t^2-3t)

I substituted 2 & 0 and the answer I got is (-2)

7. Originally Posted by katchat64
$\int_0^2{(2t-3)\,dx}$

What I got for the antiderivative of (2t-3) is (t^2-3t)

I substituted 2 & 0 and the answer I got is (-2)

Your antiderivative $F(x) = t^2 - 3t$.

$F(2) = 2^2 - 3(2) = 4 - 6 = -2$.

$F(0) = 0^2 - 3(0) = 0$

$F(2) - F(0) = -2 - 0 = -2$.

So yes it's correct

8. Thanks so much!