Question

integrate x(3+x^2)^1/2 between limits of 1 and 0
(x times square root of 3 plus x squared)

Answer 1

\[ \int\limits_{0}^{1} x \sqrt{3+x^2} dx\]

Answer 2

right?

Answer 3

yes! :)

Answer 4

i've tried by parts and got 8/3 for first bit but not sure how to work out the second bit

Answer 5

\[(1/2) \int\limits_{o}^{1} (2x)(x+x^2)^{1/2} dx\]

Answer 6

oho there is 3 inplace of first x :)

Answer 7

\[2/3 [(3+x^2)^{3/2}]\]

Answer 8

now u can put the limits

Answer 9

(2/3)(3+1)^(3/2)

Answer 10

(2/3)[(3+1)^(3/2)-(3)^(3/2)]

Answer 11

the answer is supposed to be \[8/3 - \sqrt{3}\]

Answer 12

but i can't get that answer after i put the limits in

Answer 13

(2/3)[8-3sqrt(3)]

Answer 14

16/3-2sqrt(3) is the answer

Answer 15

is it wrong?

Answer 16

\[\int\limits _0^1x \sqrt{\left(3+x^2\right)}dx = \frac{8}{3}-\sqrt{3} \]