Question

use the limit definition of definite integral to evalute int_{0}^{1} (2x+3)dx
help me please

Answer 1

\[\int\limits_{0}^{1} (2x+3)dx\]Like this, right?

Answer 2

yes

Answer 3

and 0 in down

Answer 4

Now we have to integrate that then install the limits 1 and 0. Agree?

Answer 5

yes

Answer 6

Okay, so integral will be:
\[[\frac{2x^2}{2}+ 3x]^1 _0\]

Answer 7

no please not in this way for rumin sum

Answer 8

Oh riemann sum.

Answer 9

yes

Answer 10

thank you for picture alredy i solve it like you and get 4 but in this case want to solve in rieman sum

Answer 11

please help me i wail waiting

Answer 12

please ????

Answer 13

\[f(x)=2x+3,a=0,b=1,\Delta x = \frac{1}{n}\]

Answer 14

\[x_i=a+i \Delta x\]compute \[f(x_i)\]and plug everything into \[\lim_{n \rightarrow +\infty}\sum_{i=1}^{n}f(x_i) \Delta x\]

Answer 15

ok then

Answer 16

i don't know what you mean for xi