Area by integration of the finite region bound by the two curves.

Homework - Q Sketch the graphs of the curves $y = 16 - x^2$ and $y = x^2 - 5x + 13$ The first thing I did was to set them equal to each other:

$16 = x^2 = x^2 - 5x + 13$

Then set that equal to $0$:

$2x^2 - 5x - 3 = 0$

Then factorise

$(2x + 1)(x - 3)=0$

Then we know $x = -\frac{1}{2}$ or $x = 3$

When $x = 3$, $y = 7$

When $x = -\frac{1}{2}$, $y = 15 \frac{3}{4}$

Then we can integrate.

$$\int_{-\frac{1}{2}}^3(16-x^2)dx=[16x - x^3/3]_{-1/2}^3=48-9=39$$

$16 x -1/2 = -8$

$-1/2^3/3 = -1/8$

$39 - (-8 -1/8) = 375/8$

• Then integrate the second equation.

$$\int_{-1/2}^3 (x^2 - 5x + 13)dx =[x^3/3 - 5x^2/2 + 13x]_{-1/2}^3=25\frac{1}{2}-(-1\frac{5}{24})=26\frac{17}{24}$$

• Then take one away from the other because we are looking for the area in the finite region bound by the curves.

(375/8) - (641/24) = 1125/24 - 641/24

= 484/24.

However I know the answer should be 14 and 7/24. I can't see where I have made the mistake. I would be very grateful for a hand. Thanks.

• Please do write mathematics with LaTeX, otherwise it is pretty difficult and cumbersome. You can follow the lead in other posts. – DonAntonio Nov 5 '13 at 19:15
• I'm guessing that means using the correct symbols, and if so, how? – Ed Prince Nov 5 '13 at 19:16
• Don't know. Some time ago there was a FAQ question with directions for LaTeX, and now that section's gone. You can try to enter other posts and click on "edit" to see the way to write mathematics with LaTeX. Basically, it is the differnece between x^2+3 and $\;x^2+3\;$ . The last one was written enclosed between dollar signs. – DonAntonio Nov 5 '13 at 19:18
• The substitution in $16x-x^3/3$ had a small mistake. I did not check the second one. By the way, it would be better to find the difference of the functions before integrating. Only one set of substitutions instead of two. – André Nicolas Nov 5 '13 at 19:20
• Was the mistake in $16x - x^3/3$ subbing -1/2 into $x^3/3$ It should be $-1/24$ which gives you the answer the that part $-191/24$ – Ed Prince Nov 5 '13 at 19:29 