# Thread: Evaluate the Integral (different equation)

1. ## Evaluate the Integral (different equation)

Is the antiderivative of $\int_2^1{(1+(z/2))\,dx}$ is (z+(z^2/2))?

2. Originally Posted by katchat64
Is the antiderivative of $\int_2^1{(1+(z/2))\,dx}$ is (z+(z^2/2))?
Almost. It's $-1 - \frac{z}{2}$.

$\int_2^1{1+\frac{z}{2}\,dx}$

$= [x+\frac{xz}{2}]^1_2$

$= 1+\frac{z}{2} - 2 - \frac{2z}{2}$

$= -1 - \frac{z}{2}$

3. Wait, you're not evaluating it right, your just taking the antiderivative of 1+ (z/2)?

4. Originally Posted by katchat64
Wait, you're not evaluating it right, your just taking the antiderivative of 1+ (z/2)?
You wrote $dx$, implying that the integrand was to be integrated with respect to x, not z.

Also, you put in limits, so your final answer should be numerical, not algebraic.

Also, are you sure the limits are the right way around? Did you mean

$\int_1^2{1+\frac{z}{2}\,dz}$ ???

If so then:

$\int_1^2{1+\frac{z}{2}\,dz}$

$\int_1^2{1\,dz}+\int_1^2{\frac{z}{2}\,dz}$

$\int_1^2{1\,dz}+\frac{1}{2}\int_1^2{z\,dz}$

$[z]^2_1+\frac{1}{2}[\frac{z^2}{2}]^2_1$

$[2-1]+\frac{1}{2}[\frac{2^2}{2} - \frac{1^2}{2}]$

$1+\frac{1}{2}[\frac{4}{2} - \frac{1}{2}]$

$1+\frac{1}{2}[\frac{3}{2}]$

$1+\frac{3}{4}$

$\frac{4}{4}+\frac{3}{4}$

$\frac{7}{4}$

5. Yes, this is what I meant... sorry about the dx... I understand it now and that's the same exact answer I got.