The value of int_1^3 sqrt{3 + x^3} ,dx lies i | vedclass

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

Question

(C) Let $I = \int_1^3 \sqrt{3 + x^3} \,dx$. Since $f(x) = \sqrt{3 + x^3}$ is an increasing function on the interval $[1, 3]$,the minimum value occurs

Vedclass · Accepted Answer

$(4, 2\sqrt{30})$

Vedclass · Answer

(C) Let $I = \int_1^3 \sqrt{3 + x^3} \,dx$. Since $f(x) = \sqrt{3 + x^3}$ is an increasing function on the interval $[1, 3]$,the minimum value occurs at $x = 1$ and the maximum value occurs at $x = 3$. $f(1) = \sqrt{3 + 1^3} = \sqrt{4} = 2$. $f(3) = \sqrt{3 + 3^3} = \sqrt{3 + 27} = \sqrt{30}$. Using the property of definite integrals,$(b - a)f(x)_{\min} \le I \le (b - a)f(x)_{\max}$. Here,$a = 1$ and $b = 3$,so $(b - a) = 2$. Thus,$2 	imes 2 \le I \le 2 	imes \sqrt{30}$. $4 \le I \le 2\sqrt{30}$. Therefore,the value of the integral lies in the interval $(4, 2\sqrt{30})$.

The value of $\int_1^3 \sqrt{3 + x^3} \,dx$ lies in the interval

Similar Questions

If $[x]$ is the greatest integer $\leq x$,then $\pi^{2} \int_{0}^{2}\left(\sin \frac{\pi x}{2}\right)(x-[x])^{[x]} d x$ is equal to :

Suppose $M = \int_{0}^{\pi / 2} \frac{\cos x}{x+2} dx$ and $N = \int_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} dx$. Then, the value of $(M - N)$ equals

$ \int_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x $

$\int_0^{\pi / 2} \frac{200 \sin x+100 \cos x}{\sin x+\cos x} d x$ is equal to (in $\pi$)

$\int_{-1}^3 \left(\tan^{-1}\left(\frac{x}{x^2+1}\right) + \tan^{-1}\left(\frac{x^2+1}{x}\right)\right) dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module