Evaluate the integral. ∫012−0d0

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Evaluate the integral.  ∫012−0d0

d2saint0 · Accepted Answer

∫012−0d0  Now, using the formula of integration as, ∫axdx=axlog⁡a+c to integrate the given integral.  ∫012−0d0=−[2−0log⁡(2)]01  =−[120log⁡(2)]01  =−121log⁡(2)+120log⁡(2)  =−12log⁡(2)+1log⁡(2)  =−1+22log⁡(2)  =12log⁡(2)  Therefore, the value of the integral is 12log⁡(2).

Evaluate the integral. ∫012−0d0

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