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

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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

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