# integration

If f(x) and g(x) are both continue fungtions such that f(g(x))=x and g(f(x) )=x,                     f(0)=0,f(3)=29. find ∫_0^3▒f(x)dx + ∫_0^29▒g(x)dx.
Find the area of a region bounded by the curve y=1/(1+x^2 ) , x positive  axis and y positive axis.


Note by Robi Fajar Bahari
5 years, 1 month ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Note that Nidhin makes the assumption that $$f(x), g(x)$$ are differentiable functions. This is not a necessary assumption (not is it valid, since we can have kinks).

Hint: The direct approach is to observe that $$f(x)$$ and $$g(x)$$ are inverse functions of each other. Since $$f(x), g(x)$$ and are continuous functions, this implies that $$f(x), g(x)$$ are strictly increasing functions.

Staff - 5 years, 1 month ago

thanks,,, thats very kind of you... :)

- 5 years, 1 month ago

thanks nidhin,, btw i'm a vocational school student and no so smart in calculus :) do you know how can i see the solutions to the previous week problems in "challenges" ?

- 5 years, 1 month ago

Hi Robi,

In the weekly challenges box, there is a tab that says "solutions" every wednesday(thursday depending where you are in the world) the solutions to all of the previous weeks problems that you looked at will be revealed.

Staff - 5 years, 1 month ago

integral 0 to 29 g(x)dx= integral 0 to 3 g(f(t))f'(t)dt,

so total integral becomes integral 0 to 3 ( f(t)dt + g(f(t)f'(t)dt) but g(f(t))= t,

so integral = integral 0 to 3 (f(t) + tf'(t)dt )= integral d(tf(t))dt= [tf(t)] from 0 to 3= 87

- 5 years, 1 month ago

This assumes that $$f(x)$$ and $$g(x)$$ are differentiable.

Staff - 5 years, 1 month ago