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Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Solution 1

From UBC Wiki

Consider the substitution u=x23. Then du=2xdx because the derivative of x23 with respect to x is 2x. Hence,

3xx23dx=322xdxx23=32duu

By the Power Rule for the integration, we get

duu=u12du=112+1u12+1+C=2u+C

where C is an arbitrary constant (the constant of integration)

So,

3xx23dx=32duu=32(2u+C)

and substituting x23 for u into the above expression now gives

3xx23dx=3u+C=3x23+C