What is the average value of sqrt [3](x) on the interval -5leqslant xleqslant 9

To find the average value of the function \( f(x) = \sqrt[3]{x} \) on the interval \([-5, 9]\), we use the formula for the average value of a continuous function over an interval \([a, b]\):<br /><br />\[<br />\text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx<br />\]<br /><br />Here, \( f(x) = \sqrt[3]{x} \), \( a = -5 \), and \( b = 9 \). We need to compute the integral:<br /><br />\[<br />\int_{-5}^{9} \sqrt[3]{x} \, dx<br />\]<br /><br />First, we find the antiderivative of \( \sqrt[3]{x} \):<br /><br />\[<br />\int \sqrt[3]{x} \, dx = \int x^{1/3} \, dx = \frac{x^{4/3}}{\frac{4}{3}} = \frac{3}{4} x^{4/3}<br />\]<br /><br />Now, we evaluate this antiderivative at the bounds \( -5 \) and \( 9 \):<br /><br />\[<br />\left. \frac{3}{4} x^{4/3} \right|_{-5}^{9} = \frac{3}{4} \left[ (9)^{4/3} - (-5)^{4/3} \right]<br />\]<br /><br />Next, we calculate \( (9)^{4/3} \) and \( (-5)^{4/3} \):<br /><br />\[<br />(9)^{4/3} = (9^{1/3})^4 = 3^4 = 81<br />\]<br /><br />\[<br />(-5)^{4/3} = \left( (-5)^{1/3} \right)^4 = (-5)^{4/3} = 5^{4/3} \cdot (-1)^{4/3} = 5^{4/3} \cdot 1 = 5^{4/3}<br />\]<br /><br />Thus,<br /><br />\[<br />\frac{3}{4} \left[ 81 - 5^{4/3} \right]<br />\]<br /><br />Now, we compute the average value:<br /><br />\[<br />\text{Average value} = \frac{1}{9 - (-5)} \cdot \frac{3}{4} \left[ 81 - 5^{4/3} \right] = \frac{1}{14} \cdot \frac{3}{4} \left[ 81 - 5^{4/3} \right] = \frac{3}{56} \left[ 81 - 5^{4/3} \right]<br />\]<br /><br />Therefore, the average value of \( \sqrt[3]{x} \) on the interval \([-5, 9]\) is:<br /><br />\[<br />\boxed{\frac{3}{56} \left[ 81 - 5^{4/3} \right]}<br />\]