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babymother [125]
2 years ago
13

Find the average value of the function on the given interval. f(x)=2 ln x; [1,e]

Mathematics
1 answer:
JulijaS [17]2 years ago
8 0

Answer:

f_{avg}=\frac{1}{e-1}

Step-by-step explanation:

We are given that a function

f(x)=2lnx

We have to find the average value of function on the given interval [1,e]

Average value of function on interval [a,b] is given by

\frac{1}{b-a}\int_{a}^{b}f(x)dx

Using the formula

f_{avg}=\frac{1}{e-1}\int_{1}^{e}lnx dx

By Parts integration formula

\int(uv)dx=u\int vdx-\int(\frac{du}{dx}\int vdx)dx

u=ln x and v=dx

Apply by parts integration

f_{avg}=\frac{1}{e-1}([xlnx]^{e}_{1}-\int_{1}^{e}(\frac{1}{x}\times xdx))

f_{avg}=\frac{1}{e-1}(elne-ln1-[x]^{e}_{1})

f_{avg}=\frac{1}{e-1}(e-0-e+1)=\frac{1}{e-1}

By using property lne=1,ln 1=0

f_{avg}=\frac{1}{e-1}

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