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I just started Calculus II and have been given the following problem:

Find the area inside the curves: $$y = 5 - 2x^2\\y = 3x^2 \\ x = -4 \\x = 4$$

I graphed the two curves on my calculator to get a rough idea of the problem but noticed that the x values aren't where the two curves cross.

I'm wondering if I should integrate on 4,-4 or if I should calculate where they cross and use those values as the minimum and maximum x values.

Thanks in advance

edit: The beggining of the problem was just about sketching the graph and drawing approximating rectangles; however the part that I was confused on was worded as such:

y = 5 - 2x^2, y = 3x^2, x = -4, x = 4

find the area S of the region

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2 Answers

up vote 4 down vote accepted

Draw the two parabolas (which meet at $x=\pm 1$), and the two lines. There are three finite regions in the picture. I think you are being asked to find their combined area. The issue is that in part of the picture, the parabola $y=5-2x^2$ is above the parabola $y=3x^2$, and in part of the picture, it is below $y=3x^2$. The simplest way to do the calculation is to treat these parts separately, then add up.

The "middle" region has $5-2x^2\ge 3x^2$. To find its area we calculate $$\int_{-1}^1 ((5-2x^2)-3x^2)\,dx.$$ The region on the "right" has $3x^2\ge 5-2x^2$. To find its area we calculate $$\int_1^4 (3x^2-(5-2x^2))\,dx.$$ The area of the region on the "left" can be calculated in a similar way. However, we might as well take advantage of symmetry. Now add up the three areas.

Remark: I would prefer to note the symmetry from the beginning, and observe that we can calculate the area to the right of the $y$-axis, then double. The area to the right of the $y$-axis is $$\int_0^1 ((5-2x^2)-3x^2)\,dx+\int_1^4 (3x^2-(5-2x^2))\,dx. $$ We could express this as a "single" integral, by noting that the area to the right of the $y$-axis is $$\int_0^4\left|(5-2x^2)-3x^2\right|\,dx.$$ However, the gain is illusory, since to deal with the absolute value sign, the best thing to do is to break up the integral at $x=1$.

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If the problem asks for the area between the two curves $y=5-2x^2$ and $y=3x^2$ over the interval from $x=-4$ to $x=4$, I would understand it to mean that you’re to find the total area between the two parabolas between the lines $x=-4$ and $x=4$. In that case you’ll have three separate areas to calculate, one between $x=-4$ and $x=-1$, one between $x=-1$ and $x=1$, and one between $x=1$ and $x=4$, and you’ll want their sum.

However, it would be helpful for us to see the exact wording.

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Thank you, I do vaguely recall my professorial mentioning this but he never worked out a problem like it. I edited the original post to include the wording from the problem –  Logan Besecker Aug 26 '12 at 18:50
    
@Logan: The problem is badly worded, but I think that my interpretation is probably the intended one. –  Brian M. Scott Aug 26 '12 at 18:52
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