# Thread: Area under a curve.

1. ## Area under a curve.

Hi folks,

I am trying to calculate the area between the curve $y = 1 / (x -1)(x - 3)$ and the line $y = - \frac{4}{3}$

The curve cuts the line at A (3/2, -4/3) and B(5/2, -4/3) and we can see the area in the attachment

So, Area, $A = \int_{\frac{3}{2}}^ {\frac{5}{2}} \frac{1}{(x - 1)(x - 3)}$

using partial fractions $\frac{1}{(x - 1)(x - 3)} = \frac{1}{(2x - 6)} - \frac{1}{(2x - 2)}$

so $A = \int_{\frac{3}{2}}^ {\frac{5}{2}} \frac{1}{(2x - 6)} dx - \int_{\frac{3}{2}}^ {\frac{5}{2}} \frac{1}{(2x - 2)} dx$

$A = \frac{1}{2} \int_{\frac{3}{2}}^ {\frac{5}{2}} \frac{2}{(2x - 6)} dx - \frac{1}{2} \int_{\frac{3}{2}}^ {\frac{5}{2}} \frac{2}{(2x - 2)} dx$

$A = \frac{1}{2} \ln | (2x - 6) | - \frac{1}{2} \ln | (2x - 2) |$

$A = [ \ln \frac {\sqrt{| 2x - 6 |}}{\sqrt{| 2x - 2 |}} ]_{\frac{3}{2}}^{\frac{5}{2}}$

$A = \ln \frac {\sqrt{| 5 - 6 |}}{\sqrt{| 5 - 2 |}} - \ln \frac {\sqrt{| 3 - 6 |}}{\sqrt{| 3 - 2 |}}$

$A = \ln \frac {1}{\sqrt{3}} - \ln \sqrt{3}$

$A = - 0.5493 - 0.5493 = - 1.0986$

The actual answer is 0.235 $unit^2$

can anyone spot the error?

2. ## Re: Area under a curve. Originally Posted by s_ingram Hi folks,

I am trying to calculate the area between the curve $y = 1 / (x -1)(x - 3)$ and the line $y = - \frac{4}{3}$

The curve cuts the line at A (3/2, -4/3) and B(5/2, -4/3) and we can see the area in the attachment

So, Area, $A = \int_{\frac{3}{2}}^ {\frac{5}{2}} \frac{1}{(x - 1)(x - 3)}$

Remember the area between two curves is $\int_a^b y_{upper}-y_{lower}~dx$. You have omitted the lower curve.
Also your calculations would be much simpler if you would write $\frac 1 {2x-6} = \frac 1 2 \cdot \frac 1 {x-3}$ and leave the $\frac 1 2$
outside your integral and similarly for the other fraction.

3. ## Re: Area under a curve.

Good point about 1/2 outside the integral. But as for the y lower, I thought that my range of integration would take care of that? If not, what area below the line y = -4/3 should be deducted?

4. ## Re: Area under a curve.

You can just add the Line -4/3 as + 4/3, and complete the integration 5. ## Re: Area under a curve. Originally Posted by s_ingram Good point about 1/2 outside the integral. But as for the y lower, I thought that my range of integration would take care of that? If not, what area below the line y = -4/3 should be deducted?
The range of integration gives the x limits. The problem is in the integrand. In your picture of the area, the upper curve is your $y = \frac 1 {(x-1)(x-3)}$ and the lower curve is $y =-\frac 4 3$. They both have to be in your integrand.

6. ## Re: Area under a curve.

If you go back 1 step of my integral, you will realize that it is just a double integral. 