Make the following substitution
to transform the integral into
.
Maple is correct.
I have calculated this problem multiple times, but never obtain the same answer that Maple provides.
The integration limits are from 1 to 4, and the integrand is e raised to the square root of x, divided by the root of x.
I have substituted "u" for the upper square root of x, which leaves me with the new integral with an integrand of e raised to the "u", divided by 2, with integration limits from 1 to 2.
When I evaluate this, the answer I obtain is approximately 2.3354. Maple, however, provides me with an answer of approximately 9.3415.
This question has me banging my head against the wall , because I feel that I am right, but technology is telling me I am wrong!
If someone could tell me what I am doing wrong, or what Maple might be doing, that would be very much appreciated. I apologize for my lack of mathematical notation. If someone could please tell me how to do the symbols on here, that would be awesome!
I'll just elaborate a bit on what was shown above.
You didn't change the LIMITS of integration when making the substitution. Notice that "Prove It" manually included the "u =..." in the limits. If you wanted to avoid making this mistake ever again, you could explicitly write "x = " or "u = " for each limit of integration. But eventually, you'll get used to this kind of thing and will succeed without such helps.
Where does the 2 out front come from? If the upper root of x is u, its derivative will be 1/2*root of x. And thus, 1/root of x will be replaced by 1/2du, and the 1/2 can be pulled outside of the integral, leaving 1/2 multiplied by the integral of e^u, evaluated from 1 to 2, right? Where am I wrong in my thinking?
By the way, how are you all doing fancy notation in the forum?
Online LaTeX Equation Editor - create, integrate and download is how I do it.
The 2 out in front (in the first line) isn't the way most people would work this; it would appear later. If you're having trouble with getting the integrand to contain du (exactly), you might need to review u-substitution.
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