Evaluate the Integral (different equation)

**katchat64** · January 20th 2009, 08:51 PM

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

**Mush** · January 20th 2009, 09:01 PM

Originally Posted by katchat64

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

Almost. It's $-1 - \frac{z}{2}$ .

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

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

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

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

**katchat64** · January 20th 2009, 09:06 PM

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

**Mush** · January 20th 2009, 09:19 PM

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:

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

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

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

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

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

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

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

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

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

$\frac{7}{4}$

**katchat64** · January 21st 2009, 05:48 PM

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

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