Mathematics

Evaluate the following integral as limit of sum:
$$\displaystyle \int_{0}^{5}(x+1)dx$$


SOLUTION
We have, $$I=\displaystyle \int_0^5 (x+1)dx$$. 

We know that
$$\displaystyle \int_a^b f(x)dx \displaystyle \lim_{h\to 0} [f(a)+f(a+h)+f(a+2h)+....+f(a+(n-1)h)]$$, where $$h=\dfrac {b-a}{n}$$

Here, $$a=0, b=5, f(x) =x+1$$ and $$h=\dfrac {5}{h}$$

$$\displaystyle \int_0^5 (x+1)dx =\displaystyle \lim_{h\to 0} h [f(0)+f(h)+f(2h)+...+f((n-1)h)] $$

$$\displaystyle \int_0^5 (x+1)dx =\displaystyle \lim_{h\to 0} h [1(1+h) +(1+2h)+(1+3h) +...+(1+(n-1)h)]$$

$$\displaystyle \int_0^5 (x+1)dx =\displaystyle \lim_{h\to 0} h [n+h (1+2+3+...+(n-1))]$$

$$\displaystyle \int_0^5 (x+1)dx =\displaystyle \lim_{h\to 0} h \dfrac {5}{n} \left [n+\dfrac {5}{n} \times \dfrac {n(n-1)}{2}\right]$$

 $$\displaystyle \int_0^5 (x+1)dx$$$$=\displaystyle \lim_{n\to \infty} \left[5+\dfrac {25}{2} \left (1+\dfrac {1}{n}\right)\right]$$

 $$\therefore I=\displaystyle \int_0^5 (x+1)dx$$$$=\dfrac {35}{2}$$

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