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I have one problem which confuses me, namely I have solve before this problem similar one. Previous problem:

Find the area of the figure bounded by $y=6x-x^2$ and $y=3x$

For solving this problem, I have set to equal these two graph to each other (find intersection points), so $6x-x^2=3x$, from this I have got $3x-x^2=0 \longrightarrow x(x-3)=0$ or $x=0$ and $x=3$, I have used Wolfram|$\alpha$ and then calculate area by this way $$\int_0^3(6x-x^2-3x)dx$$

When I calculated this one, I have got $4.5$, which is correct answer, because the book has this answer. The next question is similar but I could not solve it:

Calculate the area of the figure bounded by $y=-x^2+6x,y=0,y=3x$

I don't understand. Are they same? What is trick of this problem? The answer is $31.5$, but I could not solve it myself. Please help me.

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Everything looks right for the first problem. The second one doesn't seem well formulated. Note that, replacing $y=3x\ $ by $y=-3x\ $ in the first problem, you get $\frac {63}2\ $ as wished : Alpha –  Raymond Manzoni Jul 4 '12 at 13:27
    
sorry how? $y=-3*x$ why? –  dato datuashvili Jul 4 '12 at 13:30
    
I merely observed that you got the fine result in this case! Another way to get the $31.5$ would be to compute the area between $y=x^2-6x\ $ and $y=3x\ $. Or to make things short : I think that the second formulation is wrong as provided! –  Raymond Manzoni Jul 4 '12 at 13:33
1  
The second problem is well-formulated. Just draw the graph of the function: the area you need to compute is included between the curves $y=0$ and $y=3x$ ($x \in [0,3]$), and between the curves $y=0$ and $y=6x-x^2$ ($x \in [3,6]$). The answer given is correct. –  D. Thomine Jul 4 '12 at 13:36
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You may compute $\displaystyle \int_0^3 3x\,dx+\int_3^6 -x^2+6x\,dx\ $ or, as proposed by Eugene, $\int_0^6 -x^2+6x\,dx\ $ minus the first area. –  Raymond Manzoni Jul 4 '12 at 14:00

1 Answer 1

up vote 7 down vote accepted

The plot which shows what's going on is :

Rendered With Mathematica

The problem as phrased the first way asks you to calculate the area above the line, below the parabola on [0,3]. The way the question is phrased the second time, requires you to find a region bordered by all three of the x-axis, the line, and the parabola. The only such region lies on [0,6], visible in the plot.

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1  
I contend that it isn't the only such region. Observe that the "above the line, below the parabola" slice from the first problem is bordered by a (measure zero) portion of the $x$ axis. The problem really should have been phrased better (though of course you've correctly interpreted it). –  Cameron Buie Jul 4 '12 at 14:28
    
@CameronBuie and in that case, the unbounded region on $[3,\infty)$ works like a charm too! –  Eugene Shvarts Jul 4 '12 at 14:31
    
True, though of course that one has infinite area. ^_^ –  Cameron Buie Jul 4 '12 at 14:39

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