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Questions:

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My attempt

a) Using this formula: $$A = \int_{a}^{b} (y_{top} - y_{bot})dx$$

Intercepts: y = y,

25 - $x^2$ = 9

0 = $x^2$ - 16

(x-4)(x+4) = 0, a = -4, b = 4

$$\int_{-4}^{4} (25-x^2-9)dx = \int_{-4}^{4} (-x^2+16)dx = -\frac{1}{3}x^3 + 16x\bigg|_{-4}^{4} = \frac{256}{3}$$ \

b) Using this formula: $$A = \int_{a}^{b} (y_{top} - y_{bot})dx$$

Intercepts: y = y,

$16-x^2 = -9$

$0 = x^2 - 25$

(x-5)(x+5), a = -5, b = 5

$$\int_{-5}^{5} (16-x^2+9)dx = \int_{-5}^{5} (-x^2+25)dx = -\frac{1}{3}x^3 + 25x\bigg|_{-5}^{5} = \frac{500}{3}$$

c) Using this formula: $$A = \int_{a}^{b} (y_{top} - y_{bot})dx$$

Intercepts: y = y,

$2x-x^2 = 2x^2 -4x$

$0 = x^2 -6x$

$0 = x(x-6)$, a = 0 b = 6

$$\int_{0}^{6} (2x-x^2-2x^2+4x)dx = \int_{0}^{6} (-3x^2+6x)dx = -\frac{3}{3}x^3 + \frac{6}{2} x^2\bigg|_{0}^{6} = -108$$

Am I doing it right?

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    I didn't check all the integrals explicitly, just one -- but your logic is good, so yes. You're doing fine. – 2017-01-29
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    Whooops, I messed up slightly on c by substracting poorly, a = 0 b = 2 answer is 4. – 2017-01-29
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    Yes, you did. Sorry. My mistake. – 2017-01-29

1 Answers 1

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I've just gone through the three questions and your answers for a & b look correct but I think you may have gone wrong on c.

My answer to questions a, b, c: Question a, b, c

For question a, I found the complete area under the curve between the points where it cuts y=9, then found the area of the rectangle underneath y=9, then subtracted them from each other.

For question b, I translated the graph up by 9 so I could integrate between the curve and the x-axis instead of y=-9.

For question c, I thought it would be easier to integrate the two graphs separately and then add them together at the end. Also, the value of the integration can't be a negative number as you can't have a negative area (to my knowledge).

Finally, you may find this website useful for graphing problems: Desmos

Ben

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    The integral can most certainly be negative. Consider the integral of $\int_{2}^{3} x(x-2)(x-3)$. It evaluates to roughly $-0.41667$ according to Wolfram. Actually, with integration, the area is negative for regions lying beneath the x-axis. Another example is the integral of $x^2-4$ on the bounds, $[-2,2]$. You are correct in saying that areas cannot be negative, however, definite integration does not give you an area, but rather a "signed" area, which may be positive or negative. In OPs case, the "-" sign makes sense because the area of the bottom region is bigger than the area of the top's. – 2017-01-29
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    Thank you, I didn't think about that. In this circumstance we do take the integral to be positive though? – 2017-01-29
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    I'm not sure which circumstance you're referring to. But consider what an integral really is -- an infinite sum. Essentially, $\int f(x)\,dx$ is equivalent to the sum, $\Sigma\,f(x)\Delta x$, except in the case of an integral, $\Delta x$ is infinitesimally small. Granted that $\Delta x$ is positive as we move along the x-axis in the positive direction, if $f(x)$ is negative, as it is in the case where f(x) lies under the x-axis in some region, then the products are all negative (i.e. (+)(-) goes to (-)), and so the sum would be negative too. It is helpful to think in those terms. – 2017-01-29
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    Thank you, I made a small mistake on getting the intercepts for c. – 2017-01-29
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    @Ben Yes, you're right. Here the integral would be positive. – 2017-01-29