The two curves $y = 3\sqrt{4-x}$ and $y = 3\sqrt{x-2}$ are given. The task is to find the area enclosed by the curves and the x-axis.

As far as I see, the root of the first function is $x = 4$ and the root of the second function is $x = 2$. Their intersection is at $x = 3$.

My question is now the following: How do I have to chose the intervals of integration such that I get the area mentioned?

Simply integrating $\int_{2}^4 \left( 3\sqrt{4-x} - 3\sqrt{x-2} \right) dx$ seems to be wrong, isn't it?


  • $\begingroup$ Area is, by definition, a positive quantity. In order to find the area between two curves, you need to determine on which interval(s) one curve is above the other, so that the difference of the two functions is positive. $\endgroup$ – Sammy Black May 12 '13 at 8:10
  • $\begingroup$ so then I would need to integrals, one from 2 to 3 where I have int(y-g) and one from 3 to 4 where I have int(g-y)? $\endgroup$ – user66280 May 12 '13 at 8:12
  • $\begingroup$ is there typo in g=3$\sqrt{x-2}$.I think it should be y=3$\sqrt{x-2}$ $\endgroup$ – iostream007 May 12 '13 at 8:28

Notice that $3 \sqrt{4 - x}$ is only defined for $x \le 4$, where it's values are non-negative. Similarly, $3 \sqrt{x - 2}$ is only defined for $x \ge 2$, where it's values are non-negative.

As you've correctly observed, the two curves intersect at $x = 3$, where $y = 3$ as well. The two curves, together with the $x$-axis (which is yet another curve, namely $y = 0$) form the boundary of a region.

Sketch the picture. (You ought to know what the graph of $y = \sqrt{x}$ looks like. Hint: it's a parabola. Now, your curves are obtained by some transformations.)

Now, the area is calculated as: $$ \begin{align} A &= \int_2^4 \min\{ 3 \sqrt{x - 2}, 3 \sqrt{4 - x} \}\;dx \\ &= \int_2^3 3 \sqrt{x - 2}\;dx + \int_3^4 3 \sqrt{4 - x}\;dx. \end{align} $$ enter image description here


As you can see here, you must integrate $$\int_2^33\sqrt{x-2}\;\mathrm{d}x + \int_3^4\sqrt{4-x}\;\mathrm{d}x$$


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