# Math Help - Help with an Intergral

1. ## Help with an Intergral

For the curve Y = -5x4 + 50x3 – 145x2 + 150x

Ive made a table for X and Y values and sketched the graph of the profile.
Then it asks me to write and expression for the area under the curve as an integral. My tutor helped me with the below answer which I just about grasp but I just wanted to know if theres anything else I need to include,

sorry, I know the integral sign isn't correct, I just used a bracket thing instead.

Area under the curve = } ab ydx
Area under the curve = } 06 (-5x4 + 50x3 – 145x2 + 150x)dx

2. Hello,
$
I=\int^{b}_{a}{-5x^4+50x^3-145x^2+150x}dx
$
than
Using
$
\int^{x}_{0}{x^n dx}= \frac{x^{n+1}}{n+1}
$

$
\int^{b}_{a}{I dx} = [-x^5+\frac{25x^4}{2}-\frac{145x^3}{3}+75x^2]^{b}_{a}
$

Now
1) Insert a in the above equation to get value v1
2) Insert b in the equation to get value v2
3) The difference in the above to values: ie; v1-v2 is your answer
..
for second put 6 and zero in place of a and b...If still in doubt feel free to ask

3. Thank you for your reply, but im sorry, im still a bit unsure, in regards to your answere, I take it that to find the area under the curve of

(integral (0-6)) (-5x4 + 50x3 – 145x2 + 150x)dx

you use the following,

$
\int^{b}_{a}{I dx} = [-x^5+\frac{25x^4}{2}-\frac{145x^3}{3}+75x^2]^{b}_{a}
$

and then you complete the above twice, once substituting once with A, then with B, and then substract the second result to obtain the area under the curve?

4. Originally Posted by bobchiba
Thank you for your reply, but im sorry, im still a bit unsure, in regards to your answere, I take it that to find the area under the curve of

(integral (0-6)) (-5x4 + 50x3 – 145x2 + 150x)dx

you use the following,

$
\int^{b}_{a}{I dx} = [-x^5+\frac{25x^4}{2}-\frac{145x^3}{3}+75x^2]^{b}_{a}
$

and then you complete the above twice, once substituting once with A, then with B, and then substract the second result to obtain the area under the curve?
Here is an example for you

Let f(x) = x

$
\int^{5}_{0} {f(x)dx}=\int ^{5}_{0}{xdx}
$

$
=[\frac{x^2}{2}]^{5}_{0}
$

$
=\frac{5^2}{2} -\frac{0^2}{2}
$

so
$