Average Function Value — Question 4

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Question 4

Find the average value of the function f(x)=x(1+x)f(x)=\sqrt{x}(1+x) on the interval [0,1][0,1].

Original worksheet page 1: question and worked solution for 6-1-004
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Question 4 - Solution

The average value of a function on [a,b][a,b] is favg=1b−a∫abf(x)dx.f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx.

Here, favg=∫01x(1+x)dx.f_{\text{avg}}=\int_{0}^{1} \sqrt{x}(1+x)\,dx.

First expand the integrand: x(1+x)=x1/2+x3/2.\sqrt{x}(1+x)=x^{1/2}+x^{3/2}.

Integrate term by term: ∫x1/2dx=23x3/2,∫x3/2dx=25x5/2.\int x^{1/2}\,dx=\frac{2}{3}x^{3/2}, \qquad \int x^{3/2}\,dx=\frac{2}{5}x^{5/2}.

Thus, ∫01x(1+x)dx=[23x3/2+25x5/2]01.\int_{0}^{1} \sqrt{x}(1+x)\,dx = \left[\frac{2}{3}x^{3/2}+\frac{2}{5}x^{5/2}\right]_{0}^{1}.

Evaluate: 23+25=10+615=1615.\frac{2}{3}+\frac{2}{5} = \frac{10+6}{15} = \frac{16}{15}.

1615\boxed{\frac{16}{15}}

Original worksheet page 2: question and worked solution for 6-1-004

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