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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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

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The integral $\displaystyle \int \dfrac {dx}{a\cos x + b\sin x}$ is of the form $\dfrac {1}{r} \ln \left [\tan \left (\dfrac {x + \alpha}{2}\right )\right ]$.
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