# The Fundamental Theorem of Calculus

• Nov 14th 2008, 02:35 PM
McDiesel
The Fundamental Theorem of Calculus
Ok, I am having a heck of a time trying to solve this one. I know that there is something to do with absolute values but not sure where I would use it. Also, do you need to graph the equation first or just set x=0? I just keep getting the wrong answer no matter how many times I have attempted to do this. Please help!

Use a definite integral to find the area between f(x)= -3+2x^2 and the x-axis over the interval [-8,1]

• Nov 14th 2008, 02:48 PM
Mathstud28
Quote:

Originally Posted by McDiesel
Ok, I am having a heck of a time trying to solve this one. I know that there is something to do with absolute values but not sure where I would use it. Also, do you need to graph the equation first or just set x=0? I just keep getting the wrong answer no matter how many times I have attempted to do this. Please help!

Use a definite integral to find the area between f(x)= -3+2x^2 and the x-axis over the interval [-8,1]

The area is given by the integral correct? But we are trying to find the area between the curve and the x-axis, which can never be negative. So what you need to compute is $A=\int_{-8}^{1}|-3+2x^2|dx$. Do you know how to do that?
• Nov 14th 2008, 03:09 PM
McDiesel
Ok, so then do I set -3+2x^=0 which gives you -3/2. Then the intervals are -3/2 to -8 and 1 to -3/2. ? If so I keep getting the wrong answer
• Nov 14th 2008, 03:11 PM
Mathstud28
Quote:

Originally Posted by McDiesel
Ok, so then do I set -3+2x^=0 which gives you -3/2. Then the intervals are -3/2 to -8 and 1 to -3/2. ? If so I keep getting the wrong answer

I believe you solved your equation wrong.
• Nov 14th 2008, 09:58 PM
McDiesel
I'm so sorry, but can you please show me. I'm not sure why I'm not getting the right answer. I have multiple questions like this that need to be answered but need an example in order to do them...
• Nov 15th 2008, 04:48 AM
skeeter
Quote:

Use a definite integral to find the area between f(x)= -3+2x^2 and the x-axis over the interval [-8,1]
$2x^2 - 3 = 0$

$2\left(x^2 - \frac{3}{2}\right) = 0$

$x = \pm \sqrt{\frac{3}{2}}$

the solution for x in the interval [-8,1] is the negative solution

$A = \int_{-8}^{-\sqrt{\frac{3}{2}}} 2x^2 - 3 \, dx - \int_{-\sqrt{\frac{3}{2}}}^1 2x^2 - 3 \, dx$

at this point, let me tell you that this is one nasty calculation of a definite integral to do by hand ... since your solution is in decimal form, I would recommend you use a graphing calculator and calculate the value using the integral provided by Mathstud.
• Nov 15th 2008, 08:12 AM
Mathstud28
Quote:

Originally Posted by McDiesel
I'm so sorry, but can you please show me. I'm not sure why I'm not getting the right answer. I have multiple questions like this that need to be answered but need an example in order to do them...

Don't ever be sorry! We are all here to help and support! The only thing we do not appreciate is those who do not try themselves, and you do not appear to do that whatsoever.
• Nov 17th 2008, 08:12 AM
McDiesel
Quick Question
How do I know which integral should be the absolute value??
• Nov 17th 2008, 12:45 PM
Mathstud28
Quote:

Originally Posted by McDiesel
How do I know which integral should be the absolute value??

What do you mean. The area between some function $f(x)$ and the x-axis on the interval $[a,b]$ is $\text{Area}=\int_a^{b}\left|f(x)\right|dx$
• Nov 17th 2008, 03:44 PM
McDiesel
Ok, here is my work

$= -3x+2x^3/3\mid_ {-8}^{-3/2} -3x+2x^3/3\mid_ {-3/2}^{1}$

$=4.5-2.25+24-341.333+4.5-2.25-3+.667=$

$= -315.166$

far from the correct answer of 324.566
• Nov 17th 2008, 03:48 PM
Mathstud28
Quote:

Originally Posted by McDiesel
Ok, here is my work

$= -3x+2x^3/3\mid_ {-8}^{-3/2} -3x+2x^3/3\mid_ {-3/2}^{1}$

$=4.5-2.25+24-341.333+4.5-2.25-3+.667=$

$= -315.166$

far from the correct answer of 324.566

You added, when it is the negative part, or the part under the x-axis you should subtract, i.e. your last two values should be negative of what they are

$\int_{-8}^{1}\left|2-3x^2\right|dx=\int_{-8}^{\frac{-3}{2}}2x^2-3dx{\color{red}-}\int_{\frac{-3}{2}}^{1}2x^2-3dx$