Answer on Question #53155 – Math – Integral Calculus
Evaluate the following integral: ∫(1−x)31dx
Solution
Let's compute the integral:
∫(1−x)31dx=−∫(1−x)3−dx=−∫(1−x)3d(−x)=−∫(1−x)−3d(1−x)=∣t=1−x∣=−∫t−3dt=−−2t−2+C==−−2(1−x)−2+C=2(1−x)21+C,
where C is an arbitrary real constant.
Answer: ∫(1−x)31dx=2(1−x)21+C .
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