What is the average value of a function # y = x^2 (x^3 + 1)^(1/2) # on the interval #[0, 2]#?

Answer 1

#26/9#

The average value of the differentiable function #f(x)# on the interval #[a,b]# can be found through
#1/(b-a)int_a^bf(x)dx#
For #f(x)=x^2(x^3+1)^(1/2)# on #[0,2]#, this yields an average value of
#1/(2-0)int_0^2x^2(x^3+1)^(1/2)dx#
Use substitution: #u=x^3+1# and #du=3x^2dx#.
#=1/6int_0^2 3x^2(x^3+1)^(1/2)dx#
Note that the bounds will change--plug the current bounds into #u=x^3+1#.
#=1/6int_1^9u^(1/2)du#
#=1/6[2/3u^(3/2)]_1^9=1/6[2/3(9)^(3/2)-2/3(1)^(3/2)]=1/6[2/3(27)-2/3(1)]#
#=1/6[54/3-2/3]=26/9#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the average value of a function ( y = x^2 \sqrt{x^3 + 1} ) on the interval ([0, 2]), you need to evaluate the definite integral of the function over that interval and then divide the result by the length of the interval.

First, compute the definite integral of the function ( y = x^2 \sqrt{x^3 + 1} ) over the interval ([0, 2]):

[ \int_{0}^{2} x^2 \sqrt{x^3 + 1} , dx ]

After finding this integral, divide the result by the length of the interval, which is (2 - 0 = 2). This will give you the average value of the function over the interval ([0, 2]).

Calculate the definite integral and then divide by 2 to find the average value.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7