Question

(q36)Find the area under the curve y = 2^2x - 3 from 0 to 2.

Answer 1

D. 1.353

Explanation:

You want the area under the curve y = 2^(2x-3) in the interval [0, 2].

Integral

The area is found by the integral ...

`\displaystyle \int_0^2{2^(2x-3)}\,dx=(1)/(8)\int_0^2{4^x}\,dx=(1)/(8ln((4)))(4^2-4^0)=(15)/(8ln((4)))\approx\boxed{1.353}`

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