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This was a question of a exam that I take today. I would like to know the area between $t^2$ and $\sqrt{t}$ when t ∈ [0,2].

My answer was $\frac{1}{6}$, since there is no common area when t>1.

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  • $\begingroup$ then why didn't you say $\dfrac13$ ? $\endgroup$ – J. W. Tanner Jul 3 at 23:29
  • $\begingroup$ What do you mean by "common area"? If one function is larger than the other, there's area between them, if its smaller, there is, too. I'd refer you to WolframAlpha to sketch both curves in to the same coordinate system to see this in action. $\endgroup$ – Viktor Glombik Jul 3 at 23:30
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Since $t^2\leq \sqrt{t}$ for $t\in [0,1]$, and $\sqrt{t}\leq t^2$ for $t\in [1,2]$; the area is given by $$\int_0^1 \big(\sqrt{t}-t^2\big)dt+\int_1^2 \big(t^2-\sqrt{t}\big)dt=\frac{1}{3}+\left(3-\frac{4\sqrt 2}{3}\right)=\frac{10-4\sqrt 2}{3}$$ For $t>1$ the area is enclosed by the two graphs and the line $x=2$.

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Since $t^2 \leq \sqrt{t}$ for $t\in [0,1]$ and, $\sqrt{t} \leq t^2$ for $t \in [1,2]$, we can't just subtract the areas of the two functions, we need to incorporate the absolute value into our integrand.

$$\displaystyle\int_0^2 |\sqrt{t}-t^2|dt$$

When we have an absolute value, we should use a piecewise function to solve the integral:

\begin{cases} \sqrt{t}-t^2 & \text{if $t\in[0,1]$} \\ -(\sqrt{t}-t^2) & \text{if $t\in[1,2]$} \end{cases}

Now we can just integrate the piecewise function above (just integrate each equation and add).

$$\displaystyle\int_0^1 \sqrt{t}-t^2dt-\displaystyle\int_1^2 \sqrt{t}-t^2dt$$

$$(\displaystyle\int_0^1 t^{\frac{1}{2}}dt-\displaystyle\int_0^1t^2dt)-(\displaystyle\int_1^2 t^{\frac{1}{2}}dt-\displaystyle\int_1^2 t^2dt)$$

$$(\frac{2}{3}t^{\frac{3}{2}}|_0^1)-(\frac{1^3-0^3}{3})-(\frac{2}{3}t^{\frac{3}{2}}|_1^2)+(\frac{2^3-1^3}{3})$$

$$\frac{10-4\sqrt{2}}{3}$$

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