Exercises 21–24 refer to \(V = C\left( {0,1} \right)\) with the inner product given by an integral, as in Example 7.

23. Compute \(\left\| f \right\|\) for \(f\) in Exercise 21.

For the pair of vectors\(\left\langle {f,g} \right\rangle \), the inner product is given by \(\left\langle {f,g} \right\rangle = \int_0^1 {f\left( t \right)g\left( t \right)dt} \).

It is given that \(f\left( t \right) = 1 - 3{t^2}\), then \(\left\langle {f,f} \right\rangle = \int_0^1 {f\left( t \right)f\left( t \right)dt} \)and\(\left\| f \right\| = \sqrt {\left\langle {f,f} \right\rangle } \).

Plug the expression for\(f\left( t \right)\)into the inner product formula\(\left\langle {f,f} \right\rangle = \int_0^1 {f\left( t \right)f\left( t \right)dt} \), as follows:

\(\begin{aligned}{}\left\langle {f,f} \right\rangle &= \int_0^1 {f\left( t \right)f\left( t \right)dt} \\ &= \int_0^1 {\left( {1 - 3{t^2}} \right)\left( {1 - 3{t^2}} \right)dt} \\ &= \int_0^1 {\left( {9{t^4} - 6{t^2} + 1} \right)\,\,dt} \\ &= \left( {\frac{{9{t^5}}}{5} - \frac{{6{t^3}}}{3} + t} \right)_0^1\\ &= \left( {\frac{9}{5} - 2 + 1 - 0} \right)\\ &= \frac{4}{5}\end{aligned}\)

Plug the expression for\(\left\langle {f,f} \right\rangle \)into\(\left\| f \right\| = \sqrt {\left\langle {f,f} \right\rangle } \)and simplify as follows:

\(\begin{aligned}{}\left\| f \right\| &= \sqrt {\left\langle {f,f} \right\rangle } \\ &= \sqrt {\frac{4}{5}} \\ &= \frac{2}{{\sqrt 5 }}\end{aligned}\)

Hence, the required value is \(\left\| f \right\| = \frac{2}{{\sqrt 5 }}\).