Hello, my teacher was in a rush because class was about to finish. She did part of the problem, and said we could finish it up at home. I have two questions. One, in the picture i circled it in red. Why does it change from 2pi to pi? My second question is all thats left is for me to plug in "2" for x and then minus "0" for x to get the answer? thanks.
Definite integral for the area of the surface
- Thread starter kidmo87
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\(\displaystyle 2\pi \int_0^2 x \sqrt{1 + \dfrac{x^2}{4}}\ dx = 2\pi \int_0^2 x \sqrt{\dfrac{x^2+4}{4}}\ dx = 2\pi \int_0^2 x * \dfrac{1}{2} * \sqrt{x^2 + 4}\ dx =\pi \int_0^2 x \sqrt{x^2 + 4}\ dx.\)
HallsofIvy
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If you are asking how he got from \(\displaystyle 2\pi\int_0^2 x\sqrt{1+ \frac{x^2}{4}}dx\) to \(\displaystyle 2\pi\int_0^2 x\sqrt{\frac{4+ x^2}{4}}dx\), it is just getting "common denominators" and adding fractions: We can think of "1" as the fraction \(\displaystyle \frac{1}{1}\), having denominator 1 while \(\displaystyle \frac{x^2}{4}\) has denominator 4. The "least common denominator" is 4: \(\displaystyle 1+ \frac{x^2}{4}= \frac{4}{4}+ \frac{x^2}{4}= \frac{x^2+ 4}{4}\).