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

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

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Other answers:

Now we have to integrate that then install the limits 1 and 0. Agree?
yes
Okay, so integral will be: \[[\frac{2x^2}{2}+ 3x]^1 _0\]
no please not in this way for rumin sum
Oh riemann sum.
yes
thank you for picture alredy i solve it like you and get 4 but in this case want to solve in rieman sum
please help me i wail waiting
please ????
\[f(x)=2x+3,a=0,b=1,\Delta x = \frac{1}{n}\]
\[x_i=a+i \Delta x\]compute \[f(x_i)\]and plug everything into \[\lim_{n \rightarrow +\infty}\sum_{i=1}^{n}f(x_i) \Delta x\]
ok then
i don't know what you mean for xi
|dw:1355579612133:dw|
so could you tell me how i will start the formula and with pluging number please
\[x_i=a+i \Delta x = 0 + i (\frac{1}{n})=i(\frac{1}{n})\] \[f(x)=2x+3\]\[f(x_i)=2(x_i)+3 = 2\left(i (\frac{1}{n})+3\right)=\frac{2}{n}i+3\]
sorry, the closing parenthesis ) should be before +3, not after +3.
\[f(x_i) \Delta x=\left(\frac{2}{n}i + 3\right)(\frac{1}{n})\]
just expand, apply the \(\lim_{n \rightarrow \infty} \sum_{i=1}^n\) on the expression, and evaluate the limit.
ok thank you

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