Question

S x/sqrt (1+x)dx |3 to 0. (Definite Integral) Integration by Tables. The answer is 8/3 according to my book, I just don't know how to get there...

Answer 1

Is the problem
\[ \int\limits_{3}^{0}\frac{x}{\sqrt{1+x}}\ dx \]

Answer 2

try u= 1+x so x= u-1 and dx = du

Answer 3

Yes, except for it's from 3 to 0. As in, 3 is on the "top" of the integral sign and the 0 is on the "bottom" of the integral sign.

Answer 4

ok, then people would say the limits go from 0 to 3
\[ \int\limits_{0}^{3}\frac{x}{\sqrt{1+x}}\ dx \]

try u= 1+x so x= u-1 and dx = du
can you do that ?

Answer 5

the new limits for u are : lower limit of u= 1+0= 1
upper limit of u = 1+3= 4

Answer 6

In other words, with u=1+x and du= d(1+x) = dx
the problem becomes
\[ \int_{1}^{4} \frac{u-1}{\sqrt{u}} \ du \]

Answer 7

So why are you adding one to the limits?

Answer 8

when you change variables you also have to change the limits

we say u= 1+x
so when x=0 (the original lower limit) u is 1
when x= 3 u is 4