# How do I evaluate an integral by interpreting it in terms of areas?

I'm really having trouble understanding this question. The definite integral is: I solved it for its areas and got -30 because the area between 7 and 9 on the x axis contains a rectangle and a triangle, the rectangle has a base of 2 and a height of twelve while the triangle also has a base of 2 but a height of 6. The area is negative due to the area being below the x-axis. However my answer is incorrect and I am at a loss as to how to correctly do this.

• wolframalpha.com/input/… – JP McCarthy Mar 18 '15 at 13:16
• Are you sure that's it? I put it into a graphing calculator and got something different – JMartinez Mar 18 '15 at 13:18
• ...that is the graph yes so the area is not below the $x$-axis. – JP McCarthy Mar 18 '15 at 13:18
• The above is the graph of $9+3x$ so indeed the area is above the $x$-axis. I imagine you are looking for the answer 66 – Dean Barber Mar 18 '15 at 13:18
• ...well it is the line of (positive) slope $3$ and $y$-intercept $9$ so will certainly be positive for $x>0$. – JP McCarthy Mar 18 '15 at 13:19

## 3 Answers

On evaluation, it yields: $$9(9) + 1.5(9^2) - 9(7) - 1.5(7^2) = 66$$ There must be something wrong with your calculation. Please do check it...

• May I ask where you got the 1.5 from? – JMartinez Mar 18 '15 at 13:20
• Also you were right I had to fix my calculator – JMartinez Mar 18 '15 at 13:22
• $\int 3x = \frac{3x^2}{2} = 1.5(x^2)$ – Kugelblitz Mar 18 '15 at 13:22
• It's okay.. no problem. Also, nicely stated question.. +1 :) – Kugelblitz Mar 18 '15 at 13:22
• thank you, for both the explanation and the compliment :) – JMartinez Mar 18 '15 at 13:23

The area is not under the x-axis because this function is positive between 7 and 9.

• The area of the rectangle is $30*2=60$ and the area of the triangle is $6*2/2=6$, so $60 + 6 = 66$ – alex14204 Mar 18 '15 at 13:23

The area in question can be taken to be that of a trapezium whose area is given

$$\frac{1}{2}(a+b)\cdot h = \frac{1}{2} (30 + 36)\cdot 2 = 66$$

Or you can interpret the figure as a rectangle and triangle. In this case you get

$$\text{Area}~~ = 2\cdot 30 + \frac{1}{2} \cdot 2 \cdot 6 = 66$$