SOLUTION 21: Compute the area of the region enclosed by the graphs of the equations $y=\tan x, y=x,$ and $y=3$ . Begin by finding the points of intersection of the two graphs. From $\displaystyle y= \tan x$ and $y=x$ we get that $$\tan x = x \ \ \longrightarrow \ \ x=0$$ Now see the given graph of the enclosed region.

Using horizontal cross-sections to describe this region, we get that $$0 \le y \le 3 \ \ and \ \ \arctan y \le x \le y \ ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{0}^{3} (Right \ - \ Left) \ dy }$$ $$= \displaystyle { \int_{0}^{3} ( y - \arctan y ) \ dy }$$ Use Integration by Parts for $\displaystyle{ \int \arctan y \ dy }$. Recall that the Integration by Parts formula is $\displaystyle{ \int u \ dv = uv - \int v \ du }$. Let $u=\arctan y$ and $dv=dy$, so that $du = \displaystyle \frac{1}{y^2+1}$ and $v=y$. Then $$\displaystyle { \int \arctan y \ dy = y \arctan y - \int y \cdot \frac{ 1 }{ y^2+1 } \ dy }$$ $$\displaystyle { = y \arctan y - \int \frac{ y }{ y^2+1 } \ dy }$$ (Now use a simple u-substitution: Let $u=y^2+1 \ \ \longrightarrow \ \ du = 2y \ dy \ \ \longrightarrow \ \ \displaystyle{ \frac{1}{2} du = y \ dy }$ . ) $$\displaystyle { = y \arctan y - \frac{1}{2} \int \frac{1}{u} \ du }$$ $$\displaystyle { = y \arctan y - \frac{1}{2} \ln|u| + C }$$ $$\displaystyle { = y \arctan y - \frac{1}{2} \ln (y^2+1) + C }$$ Continuing with the definite integral we get that $$\displaystyle { \int_{0}^{3} ( y - \arctan y ) \ dy = \Big( \frac{y^2}{2} - \Big( y \arctan y - \frac{1}{2} \ln(y^2+1) \Big) \Big) \Big\vert_{0}^{3} }$$ $$= \displaystyle{ \Big( \frac{y^2}{2} - y \arctan y + \frac{1}{2} \ln(y^2+1) \Big) \Big\vert_{0}^{3} }$$ $$= \displaystyle{ \Big( \frac{(3)^2}{2} - (3) \arctan (3) + \frac{1}{2} \ln((3)^2+1) \Big) } - \Big( \frac{(0)^2}{2} - (0) \arctan (0) + \frac{1}{2} \ln((0)^2+1) \Big)$$ $$= \displaystyle{ \frac{9}{2} - 3 \arctan 3 + \frac{1}{2} \ln 10 - (0) + (0) - \frac{1}{2} \ln 1 \Big) }$$ $$= \displaystyle{ \frac{9}{2} - 3 \arctan 3 + \frac{1}{2} \ln 10 - \frac{1}{2}(0) }$$ $$= \displaystyle{ \frac{9}{2} - 3 \arctan 3 + \frac{1}{2} \ln 10 }$$