# Math Help - Beginner at definite integrals

1. ## Beginner at definite integrals

Could someone show me how to evaluate this: http://www.wolframalpha.com/input/?i=definite+integral+(x^2+%2B+2x+-+5)+dx%2C+x+%3D+1..4 please? (using right endpoints which I guess is the "default?")

Wolfram gives me the final answer but it does not show me the precise steps. I did the work myself but I am not at home so I can't scan my work and plus my internet doesn't work at home so I can't send it either way but I got 7 as the final answer but it's supposed to be 21 according to my teacher and wolfram. Please help!

Any help would be greatly appreciated!

2. First, determine what the integral is for $f(x) = x^2 +2x -5$:

$
F(x) = \int f(x) dx = \int (x^2 +2x -5) dx = \frac {x^3} 3 + x^2 -5x + C
$

Did you get that? Note that I used a lower-case f to denote the original function, and capital F to denote the integral. Next, evaluate F(x) at the point x=4, and subtract the value at x = 1:

$
F(4) - F(1) = (\frac {4^3} 3 + 4^2 - 5 \cdot 4 + C) - (\frac {1^3} 3 + 1^2 -5 \cdot 1 - C) = \frac {64 - 1} 3 + 16 - 20 - 1 + 5 = 21
$

Note that the constant C cancels out. Since the constant of integration always cancels when evaluating definite integrals, most people simply don't bother including it in their work, but I think it's good practice to not forget about it. Hope this helps!

3. Just to point this out.. you can break that function down between terms...

For f(x) = x2 + 2x -5 you can do this to integrate

$\int x^2 + \int 2x - \int 5$

to do this you have a simple formula of $\int x^n = x^{n+1} / n + 1$

in the first case you have x^2.... following the formula this results in (x^3) / (3)

in the case of constants such as the last 5 in the expression you simply tack on a x because when you derive 5x you get 5

This way you can break it down and say ok I have three different indiviual integrals to break down and then you can sum up the integrals afterwards.

Then since this is a definite integral you will get an number not a function.

when you have f(x) properly integrated then you can go ahead and take the upperbound in this case of 4 and sub it in for x

then you take the lower bound and sub it in for x and then you simply take your upperbound result and subtract from it your lower bound result

hope this helps