# 5 - Applications of integrations Questions Answers

y=px+a/p

Asked By: FRANCISBHORGIA 7 year ago

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 7 year ago

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 I1 + I2 + I3 + I+ I5 = 450, where In = n.0òn f(x) dx. Find f(x).

Asked By: KAMAL 8 year ago

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

0òn f(x) dx

consider x = ny so this integration will become 0ò1 f(ny) ndy = n0ò1 f(y) dy = n2 0ò1 f(x) dx

now by using these conditions I1 = 13 0ò1 f(x) dx similarly others

put these values and get the answer

Show that 0òp q3 ln sin q dq = 3p/2 0òp q2 ln [Ö2 sin q] dq.

Asked By: KAMAL 8 year ago

let I = 0òp q ln sin q dq

on aplying property of definite integral

I =  0òp (π-q) ln sin q dq

so 2 I =  0òp π ln sin q dq

or I = π/2  0ò ln sin q dq

or I = 2π/2  0òp/2 ln sin q dq    this is due to property

similarly solve  0òp q2 ln sin q dq  and   then 0òp q3 ln sin q d

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

Let g(x) be a continuous function such that    0ò1 g(t) dt = 2.   Let f(x) = 1/2 0òx (x-t)2 g(t) dt then find f '(x) and hence evaluate f "(x).

Asked By: KAMAL 8 year ago

f(x) = 1/2 0òx (x-t)2 g(t) dt

or f(x) = 1/2 0òx2 g(t) dt + 1/2 0òt2 g(t) dt - 0òx t g(t) dt

or f(x) = 1/2 x2 0ò g(t) dt + 1/2 0òt2 g(t) dt - x 0ò 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)