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
$$ \displaystyle { \int_{0}^{3} ( y - \arctan y ) \ dy
= \Big( \frac{y^2}{2} - \Big( y \arctan y - \frac{1}{2} \ln(y^2+1) \Big) \Big) \Big\vert_{0}^{3} } $$
$$ = \displaystyle{ \Big( \frac{y^2}{2} - y \arctan y + \frac{1}{2} \ln(y^2+1) \Big) \Big\vert_{0}^{3} } $$
$$ = \displaystyle{ \Big( \frac{(3)^2}{2} - (3) \arctan (3) + \frac{1}{2} \ln((3)^2+1) \Big) }
- \Big( \frac{(0)^2}{2} - (0) \arctan (0) + \frac{1}{2} \ln((0)^2+1) \Big) $$
$$ = \displaystyle{ \frac{9}{2} - 3 \arctan 3 + \frac{1}{2} \ln 10
- (0) + (0) - \frac{1}{2} \ln 1 \Big) } $$
$$ = \displaystyle{ \frac{9}{2} - 3 \arctan 3 + \frac{1}{2} \ln 10 - \frac{1}{2}(0) } $$
$$ = \displaystyle{ \frac{9}{2} - 3 \arctan 3 + \frac{1}{2} \ln 10 } $$
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