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Problem 3 Medium Difficulty

Use the indicated entry in the Table of Integrals on the Reference Pages to evaluate the integral.

$ \displaystyle \int_1^2 \sqrt{4x^2 - 3}\ dx $ ; entry 39


$\sqrt{13}-\frac{3}{4} \ln (4+\sqrt{13})-\frac{1}{2}+\frac{3}{4} \ln 3$


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Video Transcript

Okay, so this question wants us to use an integral table to find the value of this expression. So to do this, it wants us to use the following formula from the table. So in order to use this, we need to somehow transform this integral into the form you squared minus a squared. Well, in this case, it's pretty easy to see that a squared must be three. But what about are you? So we want to make you squared equal to four x squared, so you must be two times X. But now that we've introduced a u substitution, we have to remember to multiply by our general factor. Sophie, do do you? We find that it's two times DX so d X equals do you over too. So now we also need to transform our limits. So you of two is equal to well, two times two, which is four. And you of one is equal to two times one which is to So now, after all that work, we have the following expression. So our inter girl is equal to the integral from 2 to 4 of the square root of you squared minus three, which is route three squared, do you? With our factor of 1/2 from the change rule. So now this is in the exact form that our table in a girl isn't. So we can just substitute. So are integral is equal to 1/2 out front times You over too square root You squared minus three minus three over two times the natural log of you plus square root of you squared minus a squared And we're evaluating this whole thing starting at two and ending at four. So if we plug in upper limit minus lower limit, we see that we get square root of 13 minus 3/4 times the natural log. So four plus squared of 13 from our upper limit, minus 1/2 minus negative. 3/4 was a plus 3/4 Ellen of three. And there really isn't much we can do to simplify this. So this becomes our final answer

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