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Sagot :
Answer:
: let's recall that the average value of a function for an interval of (a,b) is given by formula: k=1b−a∫baf(x)dx where; k:average value k=16−2∫62(x2−2x+5)dx k=14(∣∣∣x33−2x22+5x∣∣∣62) k=14(∣∣∣x33−x2+5x∣∣∣62) k=14[(633−62+5⋅6)−(233−22+5⋅2)] k=14[(2163−36+30)−(83−4+10)] k=14[(2163−6)−(83+6)] k=14[216−183−8+183] k=14[1983−263] k=14[1723] k=17212
QuestionThe average of a function over an interval is computed as (1/width of interval) times the definite integral of the function evaluated over the interval. The indefinite integral of e^2x is (1/2)e^2x. So the answer is found by evaluating:(1/2)*[(1/2)[e^8 - e^4]], or (1/4)[e^8 - e^4]which equals about 731.6.
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Brainly.com
Question
Find the average value of f(x)=2/x over the interval [1, 3].
Answer · 0 votes
[tex] \bf slope = m = \cfrac{rise}{run} \implies \cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby \begin{array}{llll}average~rate\of~change\end{array}\\-------------------------------\\f(x)= \cfrac{2}{x} \qquad \begin{cases}x_1=1\x_2=3\end{cases}\implies \cfrac{f(3)-f(1)}{3-1}\implies \cfrac{\quad \frac{2}{3}-\frac{2}{1}\quad }{2}\\\\cfrac{\quad \frac{2-6}{3}\quad }{2}\implies \cfrac{\quad \frac{-4}{3}\quad }{\frac{2}{1}}\implies \cfrac{-4}{3}\cdot \cfrac{1}{2}\implies -\cfrac{2}{3} [/tex]
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Wyzant
Question
Find the average value of the function f(x)=x^3 over the interval [0,2] and find the value(s) of x at which the function assumes the average values
Answer · 0 votes
The average value of f is defined as: 1/(b-a)∫ f(x) dx (where integral is evaluated from a to b) If we are to integrate f(x) = x3 we get: (1/4)* (x4) Applying formula for average value: [1/(b-a)]*[(1/4)*(x4)]a to b Evaluating this result where a = 0 and b = 2: [1/(2-0)]*[(1/4)*(x4)]a to b =(1/2)*[((1/4)*x4)]a to b =(1/2)*[((1/4)*(2)4) - (2*(0^4))] =(1/2)*[((1/4)*16)-0] =(1/2)*(4) =2
The average rate of change over the interval [a,b], or the secant line between the points a and b on the function f(x), is [f(a) - f(b)]/[a-b]. So, substitute a for 1 and b for 5, and you get [f(1) - f(5)]/[1–5]. The quotient of that is your average rate of change.
the average value of f(x) on [a,b] is ∫[a,b] f(x) dx ----------------------- b-a f' = 3x^2-6x f = x^3-3x^2+4 so, you want ∫[-1,3] x^3-3x^2+4 dx -------------------------- 3 - (-1) which I'm sure you can do.
1/2 e 2 - 1/2 or 3.19 Given: f(x)=2x 2 e 2x [0, 1] The average value of a function is: Where: a and b -intervals [a,b] f(x) - given function Substitute the values to the formula: In the integration of the function, we will use integration by parts: Let: u = 2x 2 dv = e 2x dx For du, get the derivative of u: du = 2(2x 2-1 ) = 4x dx For v, integrate dv: v = 1/2 e 2x Substitute the values to the integration by parts formula, and plug it in the solution: Get the integration by parts of xe 2x dx and let: u = x dv = e 2x dx for du,get the derivative of x du = dx For v, integrate dv v = 1/2 e 2x Substitute the values to the integration by parts formula, and plug it in the solution: Final answer: The average value of the function is 1/2 e 2 - 1/2 or 3.19
Step-by-step explanation:
plz brian list Oh and the real answer is k=17212
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