Question

See equation: integration with a square root

Answer 1

\[\int\limits_{0}^{1}\sqrt{x}(x+1) dx = \]

Answer 2

http://www.wolframalpha.com/input/?i=integral+of+sqrx*%28x%2B1%29dx

Answer 3

Distribute
\[=\int\limits_{}^{}x \sqrt{x} + \sqrt{x} dx\]

Answer 4

I'm not allowed to use any graphing utilities or calcs

Answer 5

\[=\int\limits_{}^{}x^{3/2} + x^{1/2} dx\]

Answer 6

think of the square root of x as x to the power of a half

Answer 7

so now your integrand is x^3/2 + x^1/2

Answer 8

I follow

Answer 9

ok now integrate that using the chain rule, you should get (2/5)*x^(5/2) + (2/3)*x^(3/2)

Answer 10

I got 16/15

Answer 11

yup, that is correct

Answer 12

even though you are not allowed to use graphing utilities or tools, you can use them to check your answers

go to wolframalpha.com and type in this:
integrate sqrt(x)*(x+1) from 0 to 1
It's a good tool to help you check if you make any mistakes

Answer 13

\[\int\limits_{0}^{1}\sqrt{x^2-2x+1}dx \] is
thanks! I have one more similar problem...

Answer 14

well for this one, you have to notice something, the expression under the square root is actually a perfect square

Answer 15

after you notice that, you should have no problem solving it, let me know if you need help

Answer 16

oooh... that makes it so much easier. I got -1/2?

Answer 17

The correct answer should be 1/2, where did you get the negative from?

Answer 18

well when I integrated I got x^2/2 - x, and I plugged in 1 and then 0 and subtracted the two values
so 1/2(1) - 1 - 0

Answer 19

ah ok, sorry there is another trick here, because the original expression was under a square root, it is defined as non-negative

Answer 20

oh. ok! thank you