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Alex73 [517]
3 years ago
5

How would I solve this

Mathematics
1 answer:
vova2212 [387]3 years ago
3 0

A=\frac{1}{2}\ln17 = 1.417

Step-by-step explanation:

The area <em>A</em> under the curve can be written as

\displaystyle A = \int_0^2\!\dfrac{4x\:dx}{1+4x^2}

To evaluate the integral, let

u = 1+4x^2 \Rightarrow du = 8xdx\:\text{or}\:\frac{1}{2}du = 4xdx

so the integral becomes

\displaystyle \int\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\int\!\dfrac{du}{u} = \dfrac{1}{2}\ln |u|

or

\displaystyle \int\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\ln |1+4x^2|

Putting in the limits of integration, our area becomes

\displaystyle A = \int_0^2\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\left.\ln |1+4x^2|\right|_0^2

\;\;\;\;= \frac{1}{2}[\ln (1+16) - \ln (1)]

\;\;\;\;=\frac{1}{2}\ln17

Note: \ln 1 = 0

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