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

1. saifoo.khan

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

2. anonymous

yes

3. anonymous

and 0 in down

4. saifoo.khan

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

5. anonymous

yes

6. saifoo.khan

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

7. anonymous

no please not in this way for rumin sum

8. anonymous

9. saifoo.khan

Oh riemann sum.

10. anonymous

yes

11. anonymous

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

12. anonymous

13. anonymous

14. anonymous

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

15. anonymous

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

16. anonymous

ok then

17. anonymous

i don't know what you mean for xi

18. anonymous

|dw:1355579612133:dw|

19. anonymous

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

20. anonymous

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

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

22. anonymous

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

23. anonymous

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

24. anonymous

ok thank you