What is the average value of a function #y=x^2-2x+4 # on the interval #[0,8]#?
By definition
so here
By signing up, you agree to our Terms of Service and Privacy Policy
To find the average value of the function ( y = x^2 - 2x + 4 ) on the interval ([0, 8]), you need to calculate the definite integral of the function over the interval ([0, 8]), and then divide that result by the length of the interval, which is (8 - 0 = 8).
The definite integral of the function ( y = x^2 - 2x + 4 ) from (0) to (8) is:
[ \int_{0}^{8} (x^2 - 2x + 4) , dx ]
[ = \left[\frac{x^3}{3} - x^2 + 4x \right]_0^8 ]
[ = \left(\frac{8^3}{3} - 8^2 + 4(8)\right) - \left(\frac{0^3}{3} - 0^2 + 4(0)\right) ]
[ = \left(\frac{512}{3} - 64 + 32\right) - (0 - 0 + 0) ]
[ = \frac{512}{3} - 32 ]
[ = \frac{512}{3} - \frac{96}{3} ]
[ = \frac{512 - 96}{3} ]
[ = \frac{416}{3} ]
Now, to find the average value, divide this integral by the length of the interval:
[ \text{Average value} = \frac{\frac{416}{3}}{8} ]
[ = \frac{416}{24} ]
[ = \frac{52}{3} ]
So, the average value of the function ( y = x^2 - 2x + 4 ) on the interval ([0, 8]) is ( \frac{52}{3} ).
By signing up, you agree to our Terms of Service and Privacy Policy
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.
- How do you use the differential equation #dy/dx=4x+(9x^2)/(3x^3+1)^(3/2)# to find the equation of the function given point (0,2)?
- What is the general solution of the differential equation #(y^2-1)xdx + (x+2)ydy = 0#?
- Explain why both y = x-3 and y = x+2 can both be considered solutions to the differential equation dy/dx - 1 = 0?
- What is a solution to the differential equation #dy/dx=2e^(x-y)# with the initial condition #y(1)=ln(2e+1)#?
- How do you find the point c in the interval #-1<=x<=1# such that f(c) is equation to the average value of #f(x)=x^2-x#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7