What is the average value of the function g(x)=[x21=x3] on the interval [0,2]?

If f(x) is a continuous function on the interval [a,b], then the average value isx―=1b−a∫abf(x)dxHere, g(x) is continuous on the interval [0,2]Therefore, the average value isx―=12−0∫02x2+1+x3dxFirst, compute the indefinite integralI=∫x21=x3dxLet u=1+x3,⇒,du=3x2dxTherefore,I=13∫u12du=13u3232=29(1=x3)32+CAnd finallyx―=12⋅29[(1+x3)32]02=19⋅((27)−(1))=269

Find out the answer to "What is the average value of the function..."

Izabelle Walsh

Answered question

2022-04-18

What is the average value of the function

g (x) = [x^{2} \sqrt{1 = x^{3}}]

on the interval [0,2]?

Answer & Explanation

xxkrnjangxxed9q

Beginner2022-04-19Added 14 answers

If f(x) is a continuous function on the interval [a,b], then the average value is

\overset{―}{x} = \frac{1}{b - a} \int_{a}^{b} f (x) d x

Here, g(x) is continuous on the interval [0,2]
Therefore, the average value is

\overset{―}{x} = \frac{1}{2 - 0} \int_{0}^{2} x^{2} + \sqrt{1 + x^{3}} d x

First, compute the indefinite integral

I = \int x^{2} \sqrt{1 = x^{3}} d x

Let

u = 1 + x^{3}, \Rightarrow, d u = 3 x^{2} d x

Therefore,

I = \frac{1}{3} \int u^{\frac{1}{2}} d u = \frac{1}{3} \frac{u^{\frac{3}{2}}}{\frac{3}{2}}

= \frac{2}{9} {(1 = x^{3})}^{\frac{3}{2}} + C

And finally

\overset{―}{x} = \frac{1}{2} \cdot \frac{2}{9} {[{(1 + x^{3})}^{\frac{3}{2}}]}_{0}^{2}

= \frac{1}{9} \cdot ((27) - (1))

= \frac{26}{9}

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 1

Didn't find what you were looking for?