Chk Answer - Find area of bounded region (integrals)

**Yogi_Bear_79** · August 26th 2006, 12:42 PM

Q. Sketch the region bounded by the graphs of: $y=9-x^2, y=5-3x$ and then find it's area.

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

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

$= -\frac{56}{6} - \left[-\frac{61}{6}\right] = -\frac{5}{6}$

**galactus** · August 26th 2006, 04:06 PM

You have some mistakes in your integration. You shouldn't have negative area.

$\int_{-1}^{4}\left[(9-x^{2})-(5-3x)\right]dx$

**Yogi_Bear_79** · August 26th 2006, 04:48 PM

I see what started me down the wrong path. How is this?

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

$=\left[\frac{1}{6}4(9-2(4))+24\right] - \left[\frac{1}{6}(-1)(9-2(-1))+24\right]$
$= \left[\frac{4}{6}+24\right]-\left[(-\frac{11}{6})+24\right]=\frac{15}{6}$

**galactus** · August 26th 2006, 05:41 PM

No, not quite. Try for 125/6

**Yogi_Bear_79** · August 26th 2006, 06:13 PM

Ok, I'm stumped, can you show me your calculations so I can compare and see what I did wrong?

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