How do you find average value of function $h (r) = \frac{3}{{(1 + r)}^{2}}$ $h(r) = 3 / (1+r)^2$ on given interval [1, 6]?

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How do you find average value of function $h (r) = \frac{3}{{(1 + r)}^{2}}$ $h(r) = 3 / (1+r)^2$ on given interval [1, 6]?

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Answer 1 · 2015-04-13T17:39:23

$\frac{3}{14}$ $3/14$

Average value would be $\frac{1}{6 - 1}$ $1/(6-1$ $\int_{1}^{6} \frac{3}{{(1 + r)}^{2}}$ $int_1^6 3/(1+r)^2$ dr. Integrate using the power rule formula to get $\frac{1}{5}$ $1/5$ (3) $[$ $[$ -1/(1+r) $]_{1}^{6}$ $]_1^6$ .

$\frac{3}{5}$ $3/5$ [- $\frac{1}{7}$ $1/7$ + $\frac{1}{2}$ $1/2$ ]= $\frac{3}{14}$ $3/14$

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How do you find average value of function $h (r) = \frac{3}{{(1 + r)}^{2}}$ $h(r) = 3 / (1+r)^2$ on given interval [1, 6]?

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How do you find average value of function h(r) = 3 / (1+r)^2h(r)=3(1+r)2h(r) = 3 / (1+r)^2 on given interval [1, 6]?

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How do you find average value of function $h (r) = \frac{3}{{(1 + r)}^{2}}$ $h(r) = 3 / (1+r)^2$ on given interval [1, 6]?