The value of the definite integral int_0^1 fr | vedclass

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Question

(A) Let $f(x) = \frac{x}{x^3 + 16}$.To find the range of $f(x)$ on $[0, 1]$,we check its derivative:$f'(x) = \frac{(x^3 + 16)(1) - x(3x^2)}{(x^3 + 16)

Vedclass · Accepted Answer

$[0, \frac{1}{17}]$

Vedclass · Answer

(A) Let $f(x) = \frac{x}{x^3 + 16}$.To find the range of $f(x)$ on $[0, 1]$,we check its derivative:$f'(x) = \frac{(x^3 + 16)(1) - x(3x^2)}{(x^3 + 16)^2} = \frac{16 - 2x^3}{(x^3 + 16)^2}$.Since $f'(x) > 0$ for all $x \in [0, 1]$,the function $f(x)$ is strictly increasing on $[0, 1]$.Thus,the minimum value is $f(0) = 0$ and the maximum value is $f(1) = \frac{1}{1+16} = \frac{1}{17}$.Using the property of definite integrals,$m(b-a) \le \int_a^b f(x) \, dx \le M(b-a)$,where $m$ and $M$ are the minimum and maximum values of $f(x)$ on $[a, b]$:$0(1-0) \le \int_0^1 \frac{x}{x^3 + 16} \, dx \le \frac{1}{17}(1-0)$.Therefore,the integral lies in the interval $[0, \frac{1}{17}]$.

The value of the definite integral $\int_0^1 \frac{x \, dx}{x^3 + 16}$ lies in the interval $[a, b]$. The smallest such interval is

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