What will be the area which is bounded by x = 0 , y = π <m

Question

What will be the area which is bounded by

x = 0, y = π / 4

and

y = \arctan (x)

?

Answer 1

The integral in the question diverges. Notice how the integrand approaches

π / 2

and

x \to \infty

. However from the comments I'm guessing you mean,

\int_{0}^{1} \arctan x d x

(as

\tan π / 4 = 1

) From here you can use integration by parts. Remember,

\frac{d (\arctan x)}{d x} = \frac{1}{1 + x^{2}}

So, with du=1,

v = \arctan x

⟹

u=x,

d v = 1 / (1 + x^{2})

,

\int_{0}^{1} \arctan x d x = [x \arctan x]_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x^{2}} d x

I'm assuming you can take it from here.

Answer 2

The area of the figure you have described is:

\int_{0}^{π / 4} \tan y d y = {[- \log | \cos y |]}_{0}^{π / 4} = \frac{1}{2} \log 2.

Answers (2)