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Evaluate by substitution. `int_0^ax(sqrt(x^2+a^2))dx`, a>0Evaluate by substitution....

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user6978788 | Student, Undergraduate | Honors

Posted February 7, 2013 at 9:18 PM via web

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Evaluate by substitution. `int_0^ax(sqrt(x^2+a^2))dx`, a>0

Evaluate by substitution. `int_0^ax(sqrt(x^2+a^2))dx` , a>0. The answer says it should be `1/3(2sqrt(2)-1)a^3` , how come?

Tagged with antiderivative, integral, math

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tiburtius | High School Teacher | (Level 3) Associate Educator

Posted February 7, 2013 at 10:18 PM (Answer #1)

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`int_0^axsqrt(x^2+a^2)dx=|(t=x^2+a^2),(dt=2xdx),(t_1=0^2+a^2=a^2),(t_2=a^2+a^2=2a^2)|=`

Our substitution is `t=x^2+a^2` and our new limits of integration are `t_1` and `t_2`.

`1/2int_(a^2)^(2a^2)sqrt t dt=1/2cdot2/3t^(3/2)|_(a^2)^(2a^2)=1/3(2^(3/2)a^3-a^3)=1/3(2sqrt2-1)a^3`` `

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