Question

evaluate the integral x^2+X+1
everything divided by x dx from 1 to e

Answer 1

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

Answer 2

u forgot that everything is divided by x

Answer 3

oh..sorry

Answer 4

it should be x^2 +z+1/x

Answer 5

x^2+1+1/x

Answer 6

if everything is divided with x then we'd have integral S(x+1+1/x)dx from 1 to e
=x^2/2+x+lnx
then replace x with e and then with 1
and (e^2/2+e+1)-(1/2+1+0)=....

Answer 7

and ur answer will be?

Answer 8

=(e^2-2e-1)/2

Answer 9

i dont get how the answer give u that what happened with the 1/2+1... so lne +1?

Answer 10

lne = 1?

Answer 11

lne=\[\log_{e} e\]

Answer 12

did you find the 'anti derivative"?

Answer 13

oh yes
\[F(x)=\frac{x^2}{2}+x+\ln(x)\]
\[F(e)=\frac{e^2}{2}+e+1\]
\[F(1)=\frac{1}{2}+1\] subtract to get
\[\frac{e^2}{2}+e-\frac{1}{2}\]