નીચેના નિશ્ચિત સંકલનનું સરવાળાના લક્ષ તરીકે મૂલ્ય શોધો:
$\int_{0}^{5}(x+1) d x$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(D) ધારો કે $I = \int_{0}^{5}(x+1) d x$.
આપણે જાણીએ છીએ કે,$\int_a^b f (x)dx = (b - a) \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(a + rh)$,જ્યાં $h = \frac{b-a}{n}$.
અહીં,$a=0, b=5$,અને $f(x)=(x+1)$.
$\Rightarrow h = \frac{5-0}{n} = \frac{5}{n}$.
$\therefore \int_0^5 (x + 1) dx = 5 \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} f\left(r \cdot \frac{5}{n}\right)$.
$= 5 \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} \left( \frac{5r}{n} + 1 \right)$.
$= 5 \lim_{n \to \infty} \frac{1}{n} \left[ \sum_{r=0}^{n-1} 1 + \frac{5}{n} \sum_{r=0}^{n-1} r \right]$.
$= 5 \lim_{n \to \infty} \frac{1}{n} \left[ n + \frac{5}{n} \cdot \frac{(n-1)n}{2} \right]$.
$= 5 \lim_{n \to \infty} \left[ 1 + \frac{5(n-1)}{2n} \right]$.
$= 5 \lim_{n \to \infty} \left[ 1 + \frac{5}{2} \left( 1 - \frac{1}{n} \right) \right]$.
$= 5 \left[ 1 + \frac{5}{2} \right] = 5 \left[ \frac{7}{2} \right] = \frac{35}{2}$.

Explore More

Similar Questions

જો $I=\int_0^{\pi / 2} \frac{d x}{5+3 \sin x}=\lambda \tan ^{-1}\left(\frac{1}{2}\right)$, તો $\lambda=$

$\int_{0}^{3} {|2 - x| \, dx}$ ની કિંમત શોધો.

$\int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \cos x} = $

$\int_0^{\pi /6} {(2 + 3{x^2})\cos 3x\,dx = } $

ધારો કે $\{x\}$ એ વાસ્તવિક સંખ્યા $x$ નો અપૂર્ણાંક ભાગ દર્શાવે છે. તો $\int_0^2 \{x\} dx = $

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo