Chk Answer - Find area of bounded region (integrals)

• Aug 26th 2006, 12:42 PM
Yogi_Bear_79
Chk Answer - Find area of bounded region (integrals)
Q. Sketch the region bounded by the graphs of: $\displaystyle y=9-x^2, y=5-3x$ and then find it's area.

A. $\displaystyle A = \int_{-1}^{4}(9-x^2)-3x dx = \left[-\frac{1}{6}x(x(2x+9)-54\right]_{-1}^{4}$

$\displaystyle = -\frac{1}{6}4(4(2*4+9)-54) - (-\frac{1}{6})-1(-1(2*(-1)+9)-54)$

$\displaystyle = -\frac{56}{6} - \left[-\frac{61}{6}\right] = -\frac{5}{6}$
• Aug 26th 2006, 04:06 PM
galactus
You have some mistakes in your integration. You shouldn't have negative area.

$\displaystyle \int_{-1}^{4}\left[(9-x^{2})-(5-3x)\right]dx$
• Aug 26th 2006, 04:48 PM
Yogi_Bear_79
I see what started me down the wrong path. How is this?

$\displaystyle \int_{-1}^{4}\left[(9-x^{2})-(5-3x)\right]dx = \left[ \frac{1}{6}x(9-2x)+24\right]_{-1}^{4}$

$\displaystyle =\left[\frac{1}{6}4(9-2(4))+24\right] - \left[\frac{1}{6}(-1)(9-2(-1))+24\right]$
$\displaystyle = \left[\frac{4}{6}+24\right]-\left[(-\frac{11}{6})+24\right]=\frac{15}{6}$
• Aug 26th 2006, 05:41 PM
galactus
No, not quite. Try for 125/6
• Aug 26th 2006, 06:13 PM
Yogi_Bear_79
Ok, I'm stumped, can you show me your calculations so I can compare and see what I did wrong?