Asked by MathisFrustrating
Evaluate integral of e^x^(1/2) / x^(1/2)
I've looked at the answer but I don't understand what people do in their steps.
When I substitute x^(1/2) for u, I get:
2du = 1/x^(1/2) dx
But what do you do with the 1/x^(1/2) dx? It just disappears in the solutions I've seen people give.
I've looked at the answer but I don't understand what people do in their steps.
When I substitute x^(1/2) for u, I get:
2du = 1/x^(1/2) dx
But what do you do with the 1/x^(1/2) dx? It just disappears in the solutions I've seen people give.
Answers
Answered by
Jai
yes. the substitution is correct:
let u = x^(1/2)
thus du = 1/[2(x^(1/2))] dx, or
dx = 2(x^(1/2)) du, or
dx = 2u du
substituting these to original integral,
integral of [e^x^(1/2) / x^(1/2)] dx
integral of [(e^u) / u] * (2u) du
the u's will cancel out:
integral of [2*e^u] du
we can readily integrate this to
2*e^u + C
substituting back the value of u,
2*e^(x^(1/2)) + C
hope this helps~ :)
let u = x^(1/2)
thus du = 1/[2(x^(1/2))] dx, or
dx = 2(x^(1/2)) du, or
dx = 2u du
substituting these to original integral,
integral of [e^x^(1/2) / x^(1/2)] dx
integral of [(e^u) / u] * (2u) du
the u's will cancel out:
integral of [2*e^u] du
we can readily integrate this to
2*e^u + C
substituting back the value of u,
2*e^(x^(1/2)) + C
hope this helps~ :)
Answered by
MathisFrustrating
This helped alot :)
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