Area between $t^2$ and $\sqrt{t}$ in interval [0,2] 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.
 A: 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 $x=2$.
A: 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).
$$\displaystyle\int_0^1 \sqrt{t}-t^2dt-\displaystyle\int_1^2 \sqrt{t}-t^2dt$$
$$(\displaystyle\int_0^1 t^{\frac{1}{2}}dt-\displaystyle\int_0^1t^2dt)-(\displaystyle\int_1^2 t^{\frac{1}{2}}dt-\displaystyle\int_1^2 t^2dt)$$
$$(\frac{2}{3}t^{\frac{3}{2}}|_0^1)-(\frac{1^3-0^3}{3})-(\frac{2}{3}t^{\frac{3}{2}}|_1^2)+(\frac{2^3-1^3}{3})$$
$$\frac{10-4\sqrt{2}}{3}$$
