# Math Help - Definite Integral

1. ## Definite Integral

I need help with definite integrals guys. The probelm asks for me to evaluate the integral wich is </^1(on top) 2(on bottom) (1+(z/2) dz Im supposed to use these 8 rules for definitive integrals to solve wich are "order of integration", "zero width interval" "additivity" amond others. And I also use these three rules wich are </^ b(on top) a(on bottom) xdx + b^2/2 -a^2/2 and theres two others for use of a c (constant) and when it is x^2. Please I need help guys, I cant seem to solve it and i got a test tomorrow.

2. 1) Remember that no one can see into your brain or read your book. You MUST learn to communicate clearly. Please work a little harder on your notation.

2) One must use 8 rules, but you supply only 3? This goes to #1. We can't read your mind. If you want all 8, you may have to list them, at least.

3) I'll take a little shot at the first piece:

$
\int_{1}^{2}(1+\frac{z}{2})dz
$

This might be the Additive Rule

$
\int_{1}^{2}1 dz+\int_{1}^{2}\frac{z}{2} dz
$

3. Sorry its hard to be clear with this stuff but i know i skipped over alot =D I understand that part, for the left side I use c(b-a) so thats -1. However for the right side if i take 1/2 as the constant, and do the same thing c(b-a) I get -.5. So -1+.-5 = -1.5 not -7/4!

Btw the 1 and 2 are switched.

4. Originally Posted by jmoh
Sorry its hard to be clear with this stuff but i know i skipped over alot =D I understand that part, for the left side I use c(b-a) so thats -1. However for the right side if i take 1/2 as the constant, and do the same thing c(b-a) I get -.5. So -1+.-5 = -1.5 not -7/4!

Btw the 1 and 2 are switched.
$\int_2^1 dz = z|_2^1 = 1 - 2 = -1$

and
$\int_2^1 \frac{z}{2} dz = \frac{1}{2} \int_2^1 z dz = \frac{1}{2} \cdot \frac{1}{2}z^2|_2^1$

$= \frac{1}{4}(1 - 4) = -\frac{3}{4}$

So you have $\int_2^1 dz + \int_2^1 \frac{z}{2} dz = -1 - \frac{3}{4} = -\frac{7}{4}$

$\int_{2}^{1}dx = \int_{2}^{\frac{3}{2}}dx + \int_{\frac{3}{2}}^{1}dx$