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hamsa

  • 2 years ago

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

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

  2. hamsa
    • 2 years ago
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    yes

  3. hamsa
    • 2 years ago
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    and 0 in down

  4. saifoo.khan
    • 2 years ago
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    Now we have to integrate that then install the limits 1 and 0. Agree?

  5. hamsa
    • 2 years ago
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    yes

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

  7. hamsa
    • 2 years ago
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    no please not in this way for rumin sum

  8. viniterranova
    • 2 years ago
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  9. saifoo.khan
    • 2 years ago
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    Oh riemann sum.

  10. hamsa
    • 2 years ago
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    yes

  11. hamsa
    • 2 years ago
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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

  12. hamsa
    • 2 years ago
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    please help me i wail waiting

  13. hamsa
    • 2 years ago
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    please ????

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

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

  16. hamsa
    • 2 years ago
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    ok then

  17. hamsa
    • 2 years ago
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    i don't know what you mean for xi

  18. sirm3d
    • 2 years ago
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    |dw:1355579612133:dw|

  19. hamsa
    • 2 years ago
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    so could you tell me how i will start the formula and with pluging number please

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

  21. sirm3d
    • 2 years ago
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    sorry, the closing parenthesis ) should be before +3, not after +3.

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

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

  24. hamsa
    • 2 years ago
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    ok thank you

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