What is the net area between $f (x) = \frac{2}{{(x + 1)}^{2}}$ $f(x) = 2/(x+1)^2$ and the x-axis over $x \in [1, 2]$ $x in [1, 2 ]$ ?

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What is the net area between $f (x) = \frac{2}{{(x + 1)}^{2}}$ $f(x) = 2/(x+1)^2$ and the x-axis over $x \in [1, 2]$ $x in [1, 2 ]$ ?

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Answer 1 · 2016-11-07T10:45:22

$\frac{1}{3}$ $1/3$
graph{(y-2/((x+1)^2))=0 [-0.168, 4.832, -0.2, 2.41]}

Explanation:

$\frac{2}{{(x + 1)}^{2}} > 0, \forall x \neq - 1$ $2/(x+1)^2>0, AAx!=-1$

$\int_{1}^{2} \frac{2}{{(x + 1)}^{2}} d x = - 2 {[\frac{1}{x + 1}]}_{1}^{2} = - 2 (\frac{1}{3} - \frac{1}{2}) = - 2 (- \frac{1}{6}) = \frac{1}{3}$ $int_1^2 2/(x+1)^2dx=-2[1/(x+1)]_1^2=-2(1/3-1/2)=-2(-1/6)=1/3$

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What is the net area between $f (x) = \frac{2}{{(x + 1)}^{2}}$ $f(x) = 2/(x+1)^2$ and the x-axis over $x \in [1, 2]$ $x in [1, 2 ]$ ?

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What is the net area between $f (x) = \frac{2}{{(x + 1)}^{2}}$ $f(x) = 2/(x+1)^2$ and the x-axis over $x \in [1, 2]$ $x in [1, 2 ]$ ?