# 5 - Applications of integrations Questions Answers

y=px+a/p

**Asked By: FRANCISBHORGIA**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

this is a special type of differential equation in which p = dy/dx

its solution will be y = cx+(a/c) here c is a constant

y-xp=x+yp

**Asked By: FRANCISBHORGIA**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

on applying p = dy/dx

(ydx-xdy)/dx = (xdx+ydy)/dx

or ydx-xdy = xdx+ydy

or dy/dx = (y-x)/(y+x)

now it is homogeneous

A function f : R® R satisfies f(x+y) = f(x) + f(y) for all x,y Î R and is continuous throughout the domain. If I_{1} + I_{2} + I_{3} + I_{4 }+ I_{5} = 450, where I_{n} = n._{0}ò^{n} f(x) dx. Find f(x).

**Asked By: KAMAL**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

according to the given condition f(nx) = n f(x)

_{0}ò^{n} f(x) dx

consider x = ny so this integration will become _{0}ò^{1} f(ny) ndy = n^{2 }_{0}ò^{1} f(y) dy = n^{2} _{0}ò^{1} f(x) dx

now by using these conditions I_{1} = 1^{3} _{0}ò^{1} f(x) dx similarly others

put these values and get the answer

Show that _{0}ò^{p} q^{3} ln sin q dq = 3p/2 _{0}ò^{p} q^{2} ln [Ö2 sin q] dq.

**Asked By: KAMAL**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

let I = _{0}ò^{p} q ln sin q dq

on aplying property of definite integral

I = _{0}ò^{p} (π-q) ln sin q dq

so 2 I = _{0}ò^{p} π ln sin q dq

or I = π/2 _{0}ò^{p } ln sin q dq

or I = 2π/2 _{0}ò^{p/2} ln sin q dq this is due to property

similarly solve _{0}ò^{p} q^{2} ln sin q dq and then _{0}ò^{p} q^{3} ln sin q dq

finally solve the right hand side of the equation to prove LHS = RHS

Let g(x) be a continuous function such that _{0}ò^{1} g(t) dt = 2. Let f(x) = 1/2 _{0}ò^{x} (x-t)^{2} g(t) dt then find f '(x) and hence evaluate f "(x).

**Asked By: KAMAL**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

f(x) = 1/2 _{0}ò^{x} (x-t)^{2} g(t) dt

or f(x) = 1/2 _{0}ò^{x }x^{2 }g(t) dt + 1/2 _{0}ò^{x }t^{2 }g(t) dt - _{0}ò^{x }x t g(t) dt

or f(x) = 1/2 x^{2} _{0}ò^{x }^{ }g(t) dt + 1/2 _{0}ò^{x }t^{2 }g(t) dt - x _{0}ò^{x } t g(t) dt

now use newton leibniz formula for middle function and product rule for first and last functions for finding f^{'} (x)

similarly get f^{''} (x)