Answer:    

GIVEN :     $f(x) = c_1\phi_1(x)+c_2\phi_2(x)+c_3\phi_3(x)$ where c1,c2,c3  constants and   $\phi_1$, $\phi_2$, $\phi_3$  are ORTHONORMAL sets on (a,b).

• Since$\phi_1$ , $\phi_2$,$\phi_3$  are ORTHONORMAL

$\int\limits_a^b [\phi{_1(x)}]^2dx=\int\limits_a^b [\phi{_2(x)}]^2dx=\int\limits_a^b [\phi{_3(x)}]^2dx=1$.......................(1)

and     $\int\limits_a^b [\phi{_m(x)}\phi{_n(x)}dx=0$   when m$\neq$n.........................(2)

SHOW THAT :   $\int\limits_{a}^b[f(x)]^2dx =c_1^2+c_2^2+c_3^2$

Consider L.H.S=   $\int\limits_a^b [f(x)]^2dx$ 

$=\int\limits_a^b [c_1\phi{_1(x)}+c_2\phi{_2(x)}+c_3\phi{_3(x)}]^2dx$   .........since it is given $f(x) = c_1\phi_1+c_2\phi_2+c_3\phi_3 $ hence substituting value of f(x)

$=\int\limits_a^b [c_1^2\phi{_1(x)}^2+ c_2^2\phi{_2(x)}^2+c_3^2\phi{_3(x)}^2+2c_1c_2\phi{_1(x)}\phi{_2(x)}+2c_1c_3\phi{_1(x)}\phi{_3(x)}+2c_2c_3\phi{_2(x)}\phi{_3(x)}]dx$ .........Expanding f(x) using (a+b+c)2= a2+b2+c2+2ab+2ac+2bc expansion

$=c_1^2\int\limits_a^b [\phi_1{(x)}^2]dx+ c_2^2\int\limits_a^b[\phi{_2(x)}^2]dx+c_3^2\int\limits_a^b[\phi{_3(x)}^2]dx+2c_1c_2\int\limits_a^b\phi{_1(x)}\phi{_2(x)}dx+2c_1c_3\int\limits_a^b\phi{_1(x)}\phi{_3(x)}dx+2c_2c_3\int\limits_a^b\phi{_2(x)}\phi{_3(x)}dx$

=  c12  + c22 +  c32    .................by (1) and (2)  

Hence we observe that 

L.H.S =  R.H.S

Hence,   $\int\limits_a^b[f(x)]^2dx =$ c12 + c22 + c32