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



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

hamsa
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yes

hamsa
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and 0 in down

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

hamsa
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yes

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

hamsa
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no please not in this way for rumin sum

viniterranova
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saifoo.khan
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Oh riemann sum.

hamsa
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yes

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

hamsa
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please help me i wail waiting

hamsa
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please ????

sirm3d
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\[f(x)=2x+3,a=0,b=1,\Delta x = \frac{1}{n}\]

sirm3d
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\[x_i=a+i \Delta x\]compute \[f(x_i)\]and plug everything into \[\lim_{n \rightarrow +\infty}\sum_{i=1}^{n}f(x_i) \Delta x\]

hamsa
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ok then

hamsa
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i don't know what you mean for xi

sirm3d
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dw:1355579612133:dw

hamsa
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so could you tell me how i will start the formula and with pluging number please

sirm3d
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\[x_i=a+i \Delta x = 0 + i (\frac{1}{n})=i(\frac{1}{n})\]
\[f(x)=2x+3\]\[f(x_i)=2(x_i)+3 = 2\left(i (\frac{1}{n})+3\right)=\frac{2}{n}i+3\]

sirm3d
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sorry, the closing parenthesis ) should be before +3, not after +3.

sirm3d
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\[f(x_i) \Delta x=\left(\frac{2}{n}i + 3\right)(\frac{1}{n})\]

sirm3d
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just expand, apply the \(\lim_{n \rightarrow \infty} \sum_{i=1}^n\) on the expression, and evaluate the limit.

hamsa
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ok thank you