1
$\begingroup$

I'm really having trouble understanding this question. The definite integral is:

enter image description here

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.

$\endgroup$
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$ – JP McCarthy Mar 18 '15 at 13:16
  • $\begingroup$ Are you sure that's it? I put it into a graphing calculator and got something different $\endgroup$ – JMartinez Mar 18 '15 at 13:18
  • $\begingroup$ ...that is the graph yes so the area is not below the $x$-axis. $\endgroup$ – JP McCarthy Mar 18 '15 at 13:18
  • $\begingroup$ 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 $\endgroup$ – Dean Barber Mar 18 '15 at 13:18
  • $\begingroup$ ...well it is the line of (positive) slope $3$ and $y$-intercept $9$ so will certainly be positive for $x>0$. $\endgroup$ – JP McCarthy Mar 18 '15 at 13:19
2
$\begingroup$

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...

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

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

$\endgroup$
  • $\begingroup$ The area of the rectangle is $30*2=60$ and the area of the triangle is $6*2/2=6$, so $60 + 6 = 66$ $\endgroup$ – alex14204 Mar 18 '15 at 13:23
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.