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I sketched out the graph to get this figure but I can't seem to find the area of the shaded region... would one Y = 4 and the other Y = 8? enter image description here

I, in all honesty, am so flabbergasted with this question, any guidance will be much appreciated. I got 28/3 for the first integral (0 to 4) and 292/3 for the second integral (4 to 16)... I feel like I'm doing this wrong.

Okay, I'm going to be honest, can someone please solve this for me? I got 320/3 as my final answer and yet it's still wrong, I'm just about given up at this point, it's literally my last question of this semester.

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4 Answers 4

up vote 1 down vote accepted

We want,

$$\left(\int_{0}^{4} \sqrt{x}\,\,dx-\int_{0}^{4}\dfrac12x\,\,dx\right)+\left(\int_{4}^{16}\dfrac12x\,\,dx-\int_{4}^{16}\sqrt{x}\,\,dx\right)$$

where the first term two terms give the area from $0$ to $4$ and the second term gives it from $4$ to $16$.

The expression becomes: $\left(\dfrac23x^{3/2}-\dfrac14x^2\right)\left.\right|_0^4+\left(\dfrac14x^2-\dfrac23x^{3/2}\right)\left.\right|_4^{16}$

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Okay, I got 16/5 for the first integral and 1072/15 for the second, after making the first 48/15, I added it to the second to get 1120/15, it's still not correct. I am utterly confused. –  Harry Johnson Dec 20 '13 at 1:13
    
@HarryJohnson According to my calculation, the area from $0$ to $4$ is $\dfrac43$ and from $4$ to $16$ is $\dfrac{68}3$, which gives a total of $24$. –  Alraxite Dec 20 '13 at 1:25
    
AHHH, I GOT IT! I'm so silly, I was inputting it all wrong, I finally understand. The answer I got was 288/12... otherwise known as 24 :))) Thank you so, so, so much! –  Harry Johnson Dec 20 '13 at 1:30

HINT Find the point of intersection, set the limits $a = 0$ and $b =$ the point you found, set up the integral with those limits and the function $f(x) - g(x)$, and set up another integral from $b$ to $16$ with $g(x) - f(x)$.

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I got 28/3 for the first integral (0 to 4) and 292/3 for the second integral (4 to 16).. but it's not correct :( –  Harry Johnson Dec 19 '13 at 20:54
    
@Harry: Can you write down your attempt at a solution in "Add another answer"? Thanks. –  Don Larynx Dec 19 '13 at 22:00
    
Oh, I apologize, I'm a novice, I have no idea where that is lol. But if it's not inconveniencing you that much, is the answer 320/3? –  Harry Johnson Dec 20 '13 at 0:22
    
@HarryJohnson: I haven't worked out the problem. The button "Add another answer" is below, near the bottom. –  Don Larynx Dec 20 '13 at 2:05

You need to set up two double integrals.

The first one:

$\int_0^4 \int_{.5x}^ \sqrt{x} dydx$

The second one:

$\int_4^{16} \int_\sqrt{x}^{.5x} dydx$

The computation is straightforward.

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1  
Nice way to do it with double integrals. –  Don Larynx Dec 19 '13 at 2:10
1  
Should the integral limits not be 0 to 4 and 4 to 16, as the point of intersection is at $x=4$? –  Guy Corrigall Dec 19 '13 at 2:22
    
@GuyCorrigall you're right! My bad, it's hard to see the tick marks without my glasses. i'll fix that. –  Lame-Ov2.0 Dec 19 '13 at 2:24
    
Okay I am still a bit confused, I don't know how to find the area? :( –  Harry Johnson Dec 19 '13 at 6:36
    
@HarryJohnson what do you mean? You find the area through integration. A double integral is the same as a single integral, just iterated once more. –  Lame-Ov2.0 Dec 19 '13 at 9:14

The area between two curves given by functions $f(x)$ and $g(x)$ is given by,

$$ \int_a^b \vert f(x)-g(x) \vert dx = \int_a^b \textbf{Top Function} - \textbf{Bottom Function }\mathrm{d}x$$

A common mistake students make when trying to compute this is that they don't change the order of the subtraction when the "Top Function" changes from $f(x)$ to $g(x)$. This happens whenever the two curves cross eachother. In your case they seem to cross at $x=4$ so that you need to evaluate an integral from $0$ to $4$ and then another one from $4$ to $16$ always making sure to have to top function come first in the subtraction.

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is it 4 to 15, or 4 to 16? –  Harry Johnson Dec 19 '13 at 20:00
    
@Harry Johnson 4 to 16, made a mistake when looking at the diagram. –  Spencer Dec 19 '13 at 20:03
    
Mr. Spencer, can you help me solve this problem? I'm just about given up at this point. I got 320/3 after adding both my integrals and yet it's still wrong. –  Harry Johnson Dec 20 '13 at 0:23
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@Harry: If you can display your work by clicking the "Add Another Answer" page at the bottom, we will be glad to help you. –  Don Larynx Dec 20 '13 at 2:05

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