# Thread: area under curve and line.

1. ## 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.

$7-x = x^2 + 1$

$0 = x^2 + x -6$

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

$x = 2 , x = -3$

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

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.

2. Hi

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

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

3. Originally Posted by running-gag
Hi

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

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?

4. I got $\frac{310}{3}$ , which is not the right answer.

Can you show me how to work it out ?

thanks.

5. 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
Originally Posted by Tweety
I got $\frac{310}{3}$ , which is not the right answer.

Can you show me how to work it out ?

thanks.
You have the functions:

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

Split $R_2$ 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 $R_2$ is:

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

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

I'll leave the rest for you. I've got $R_2 = \dfrac{103}6$

6. Area of R2 = $\int_{0}^2 [ x^2 + 1] dx + \int_{2}^7 [ 7 - x] dx$

7. Thanks I get it now.