# Trig and Integration help please?

## Find, to one decimal place, the area bounded by $x = 0$, $y = 0$, $y = \cos x$ and $y = {\cos}^{-} 1 x$. Thanks!

##### 1 Answer
Oct 4, 2017

${\int}_{0}^{0.74} \cos \left(x\right) d \setminus x + {\int}_{0.74}^{1} \arccos \left(x\right) d \setminus x = 0.8$

#### Explanation:

The plot of $\cos \left(x\right)$ in purple and ${\cos}^{- 1} \left(x\right)$ in blue

The coordinate of intersection $\left(x , y\right) = \left(0.74 , 0.74\right)$
So, between $\left(0 , 0.74\right)$,$\cos \left(x\right)$ is the bounding function on the upper part, while between $\left(0.74 , 1\right)$, ${\cos}^{- 1} \left(x\right)$ is the bounding function.
Therefore the area of interest is:
${\int}_{0}^{0.74} \cos \left(x\right) d \setminus x + {\int}_{0.74}^{1} \arccos \left(x\right) d \setminus x = 0.8$