10.0 Problems on Integral Functions

Let function $f\left( x \right)$

In this type of problem, a equation is given which contain $f$ inside an integral term and also as a independent term in the equation.

For Example, $$f\left( x \right) = {x^2} + \int\limits_0^{{\pi \over 2}} {\cos \left( {x + y} \right) \cdot f\left( y \right)dy} $$

Take the above example and expand ${\cos \left( {x + y} \right)}$

$$f\left( x \right) = {x^2} + \cos x\int\limits_0^{{\pi \over 2}} {\cos y \cdot f\left( y \right)dy} - \sin x\int\limits_0^{{\pi \over 2}} {\sin y \cdot f\left( y \right)dy} $$

Let $P = \int\limits_0^{{\pi \over 2}} {\cos y \cdot f\left( y \right)dy} $ and $Q = \int\limits_0^{{\pi \over 2}} {\sin y \cdot f\left( y \right)dy} $

Then equation become $$f\left( x \right) = {x^2} + P\cos x - Q\sin x$$

Now solve for P and Q by using above value of $f\left( x \right)$

$$P = \int\limits_0^{{\pi \over 2}} {\cos y\left( {{y^2} + P{\mathop{\rm cosy}\nolimits} - Q{\mathop{\rm siny}\nolimits} } \right)dy} $$

$$Q = \int\limits_0^{{\pi \over 2}} {\sin y\left( {{y^2} + P{\mathop{\rm cosy}\nolimits} - Q{\mathop{\rm siny}\nolimits} } \right)dy} $$

By solving $$P = \int\limits_0^{{\pi \over 2}} {{y^2}\cos ydy} + P\int\limits_0^{{\pi \over 2}} {{{\cos }^2}ydy} - Q\int\limits_0^{{\pi \over 2}} {\cos y\sin ydy} $$

$$Q = \int\limits_0^{{\pi \over 2}} {{y^2}\sin ydy} + P\int\limits_0^{{\pi \over 2}} {\cos y\sin ydy} - Q\int\limits_0^{{\pi \over 2}} {{{\sin }^2}ydy} $$

Let $${I_1} = \int\limits_0^{{\pi \over 2}} {{y^2}\cos ydy} $$ $${I_2} = \int\limits_0^{{\pi \over 2}} {{{\cos }^2}ydy} $$ $${I_3} = \int\limits_0^{{\pi \over 2}} {\cos y\sin ydy} $$ $${I_4} = \int\limits_0^{{\pi \over 2}} {{y^2}\sin ydy} $$ $${I_5} = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}ydy} $$

Solve for integrals and put the values in above equations to get the values of $P$ and $Q$.

$${I_1} = \int\limits_0^{{\pi \over 2}} {y\cos ydy} $$ $${I_1} = \left[ {y\int {\cos ydy} } \right]_0^{{\pi \over 2}} - \int\limits_0^{{\pi \over 2}} {\left[ {{{dy} \over {dy}}\int {\cos ydy} } \right]dy} $$ $${I_1} = \left[ {y\sin y} \right]_0^{{\pi \over 2}} - \int\limits_0^{{\pi \over 2}} {\sin ydy} $$ $${I_1} = \left[ {y\sin y} \right]_0^{{\pi \over 2}} + \left[ {\cos y} \right]_0^{{\pi \over 2}}$$ $${I_1} = {\pi \over 2} - 1$$

$${I_2} = \int\limits_0^{{\pi \over 2}} {{{\cos }^2}ydy} $$ $${I_2} = \int\limits_0^{{\pi \over 2}} {{{\cos 2y + 1} \over 2}dy} $$ $${I_2} = \left[ {{{\sin 2y} \over 4} + {y \over 2}} \right]_0^{{\pi \over 2}}$$ $${I_2} = {\pi \over 4}$$

$${I_3} = \int\limits_0^{{\pi \over 2}} {\cos y\sin ydy} $$ $${I_3} = {1 \over 2}\int\limits_0^{{\pi \over 2}} {\cos 2ydy} $$ $${I_3} = {1 \over 4}\left[ {\sin 2y} \right]_0^{{\pi \over 2}} = 0$$

$${I_4} = \int\limits_0^{{\pi \over 2}} {y\sin ydy} $$ $${I_4} = \left[ {y\int {\sin ydy} } \right]_0^{{\pi \over 2}} - \int\limits_0^{{\pi \over 2}} {\left[ {{{dy} \over {dy}}\int {\sin ydy} } \right]dy} $$ $${I_4} = \left[ { - y{\mathop{\rm co}\nolimits} {\mathop{\rm s}\nolimits} y} \right]_0^{{\pi \over 2}} + \int\limits_0^{{\pi \over 2}} {co{\mathop{\rm s}\nolimits} ydy} $$ $${I_4} = \left[ { - y\cos y} \right]_0^{{\pi \over 2}} + \left[ {\sin y} \right]_0^{{\pi \over 2}}$$ $${I_4} = 1$$

$${I_5} = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}ydy} $$ $${I_5} = \int\limits_0^{{\pi \over 2}} {1 - {{\cos }^2}ydy} $$ $${I_5} = \int\limits_0^{{\pi \over 2}} {1dy} - {I_2}$$ $${I_5} = {\pi \over 2} - {\pi \over 4} = {\pi \over 4}$$ Now above equations become $$P = {\pi \over 2} - 1 + {\pi \over 4}P$$ $$Q = 1 - {\pi \over 4}Q$$ Solving these we get $$P = {{2\pi - 4} \over {4 - \pi }}$$ $$Q = {4 \over {4 + \pi }}$$ Put these values in the $f\left( x \right)$ we get $$f\left( x \right) = x + \left( {{{2\pi - 4} \over {4 - \pi }}} \right)\cos x - \left( {{4 \over {4 + \pi }}} \right)\sin x$$

Question 24. Let two functions $f\left( x \right)$ and $g\left( x \right)$ be defined as $R \to R$ such that $$f\left( x \right) = {{{x^3}} \over 2} + 1 - x\int\limits_0^x {g\left( t \right)dt} $$ $$g\left( x \right) = x - \int\limits_0^1 {f\left( t \right)dt} $$ Then find $f\left( x \right)$ and $g\left( x \right)$.

Solution: Let $\int\limits_0^1 {f\left( t \right)dt} = A = {\rm{constant}}$ then $$g\left( x \right) = x - A$$ and $$f\left( x \right) = {{{x^3}} \over 2} + 1 - x\int\limits_0^x {\left( {t - A} \right)dt} $$ $$f\left( x \right) = {{{x^3}} \over 2} + 1 - x\left[ {{{{t^2}} \over 2} - At} \right]_0^x$$ $$f\left( x \right) = {{{x^3}} \over 2} + 1 - x\left[ {{{{x^2}} \over 2} - Ax} \right]$$ $$f\left( x \right) = 1 + A{x^2}$$ Now $$A = \int\limits_0^1 {\left( {1 + A{t^2}} \right)dt} $$ $$A = \left[ {t + A{{{t^3}} \over 3}} \right]_0^1$$ $$A = 1 + {A \over 3}$$ $$A = {3 \over 2}$$ Using the value of A, we get $$f\left( x \right) = 1 + {3 \over 2}{x^2}$$ $$g\left( x \right) = x - {3 \over 2}$$