# area under curve and line.

• Apr 24th 2009, 08:26 AM
Tweety
area under curve and line.
The diagram shows a sketch of part of the curve with equation y=x2+1 and the line with equation y=7−x. The finite region R1 is bounded by the line and the curve. The finite region R2 is below the curve and the line and is bounded by the positive x- and y-axes as shown in the diagram.

(a) Find the area of R1.
(b) Find the area of R2.

$\displaystyle 7-x = x^2 + 1$

$\displaystyle 0 = x^2 + x -6$

$\displaystyle 0 = (x+3) (x-2)$

$\displaystyle x = 2 , x = -3$

Area of R1 = $\displaystyle \int_{-3}^2 [ 7-x-x^2 + 1] dx$

I can work this out, but How would I go about working the area of region 2? Becasue there's also the area of region 1 inside region 2.
• Apr 24th 2009, 08:35 AM
running-gag
Hi

Area of R1 = $\displaystyle \int_{-3}^2 [ 7-x-x^2 - 1] dx$

For R2, you need to split the integral into 2 parts : from 0 to 2 and from 2 to 7
• Apr 24th 2009, 08:50 AM
Tweety
Quote:

Originally Posted by running-gag
Hi

Area of R1 = $\displaystyle \int_{-3}^2 [ 7-x-x^2 - 1] dx$

For R2, you need to split the integral into 2 parts : from 0 to 2 and from 2 to 7

So would I minus the integrals after ? To get the area of R2?
• Apr 24th 2009, 09:10 AM
Tweety
I got $\displaystyle \frac{310}{3}$ , which is not the right answer.

Can you show me how to work it out ?

thanks.
• Apr 24th 2009, 10:45 AM
earboth
Quote:

Originally Posted by running-gag
...
For R2, you need to split the integral into 2 parts : from 0 to 2 and from 2 to 7

Quote:

Originally Posted by Tweety
I got $\displaystyle \frac{310}{3}$ , which is not the right answer.

Can you show me how to work it out ?

thanks.

You have the functions:

$\displaystyle f(x)=x^2+1$ and $\displaystyle g(x)=7-x$

Split $\displaystyle R_2$ into two parts: One area is completely below the graph of the quadratic function, the second one is completely below the straight line. According to running-gag's post the area $\displaystyle R_2$ is:

$\displaystyle R_2=\int_0^2 f(x) dx +\int_2^7 g(x) dx$

$\displaystyle R_2=\int_0^2 \left(x^2+1 \right)dx +\int_2^7 (7-x) dx$

I'll leave the rest for you. I've got $\displaystyle R_2 = \dfrac{103}6$
• Apr 24th 2009, 10:46 AM
running-gag
Area of R2 = $\displaystyle \int_{0}^2 [ x^2 + 1] dx + \int_{2}^7 [ 7 - x] dx$
• Apr 24th 2009, 11:00 AM
Tweety
Thanks I get it now.