Evaluate int_2^3 x^2 dx as the limit of a sum | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We use the definition of the definite integral as the limit of a sum: $\int_a^b f(x) dx = \lim_{h 	o 0} h \sum_{r=0}^{n-1} f(a+rh)$,where $nh = b

Vedclass · Accepted Answer

$\frac{19}{3}$

Vedclass · Answer

(D) We use the definition of the definite integral as the limit of a sum: $\int_a^b f(x) dx = \lim_{h 	o 0} h \sum_{r=0}^{n-1} f(a+rh)$,where $nh = b-a$.Here,$a=2$,$b=3$,and $f(x)=x^2$. Thus,$nh = 3-2 = 1$.$I = \lim_{h 	o 0} h \sum_{r=0}^{n-1} (2+rh)^2 = \lim_{h 	o 0} h \sum_{r=0}^{n-1} (4 + 4rh + r^2h^2)$.$I = \lim_{h 	o 0} [4nh + 4h^2 \sum_{r=0}^{n-1} r + h^3 \sum_{r=0}^{n-1} r^2]$.Using the formulas $\sum_{r=0}^{n-1} r = \frac{(n-1)n}{2}$ and $\sum_{r=0}^{n-1} r^2 = \frac{(n-1)n(2n-1)}{6}$:$I = \lim_{h 	o 0} [4nh + 4h^2 \frac{(n-1)n}{2} + h^3 \frac{(n-1)n(2n-1)}{6}]$.Since $nh=1$,we have $I = \lim_{h 	o 0} [4(1) + 2(nh-h)(nh) + \frac{(nh-h)(nh)(2nh-h)}{6}]$.Substituting $nh=1$ and $h 	o 0$:$I = 4 + 2(1)(1) + \frac{(1)(1)(2)}{6} = 4 + 2 + \frac{1}{3} = 6 + \frac{1}{3} = \frac{19}{3}$.

Evaluate $\int_2^3 x^2 dx$ as the limit of a sum.

Similar Questions

The value of $\lim_{n \to \infty} \left[ \frac{1}{n}\sin \left( \frac{1}{n} \right)\left( \cos \left( \frac{1}{n} \right) \right)^2 + \frac{1}{n}\sin \left( \frac{2}{n} \right)\left( \cos \left( \frac{2}{n} \right) \right)^2 + \dots + \frac{1}{n}(\sin 1)(\cos 1)^2 \right]$ is

$\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k}{n^2+k^2} = $

The value of $\lim _{n \rightarrow \infty} \left\{ \frac{\sqrt{n+1}+\sqrt{n+2}+\ldots+\sqrt{2n-1}}{n^{3/2}} \right\}$ is

If $a = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{2n}{n^2+k^2}$ and $f(x) = \sqrt{\frac{1-\cos x}{1+\cos x}}$,$x \in (0, 1)$,then:

$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{2 n} \frac{r}{\sqrt{n^2+r^2}}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module