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

1. saifoo.khan

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

2. hamsa

yes

3. hamsa

and 0 in down

4. saifoo.khan

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

5. hamsa

yes

6. saifoo.khan

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

7. hamsa

no please not in this way for rumin sum

8. viniterranova

9. saifoo.khan

Oh riemann sum.

10. hamsa

yes

11. hamsa

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

13. hamsa

14. sirm3d

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

15. sirm3d

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

16. hamsa

ok then

17. hamsa

i don't know what you mean for xi

18. sirm3d

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19. hamsa

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

20. sirm3d

$x_i=a+i \Delta x = 0 + i (\frac{1}{n})=i(\frac{1}{n})$ $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

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

22. sirm3d

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

23. sirm3d

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

24. hamsa

ok thank you