What is the average value of sin(x) on the interval -6leqslant xleqslant -1 ?

To find the average value of \( \sin(x) \) on the interval \([-6, -1]\), we use the formula for the average value of a continuous function \( f(x) \) over an interval \([a, b]\):<br /><br />\[<br />\text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx<br />\]<br /><br />In this case, \( f(x) = \sin(x) \), \( a = -6 \), and \( b = -1 \). So, we need to compute:<br /><br />\[<br />\text{Average value} = \frac{1}{-1 - (-6)} \int_{-6}^{-1} \sin(x) \, dx<br />\]<br /><br />Simplify the denominator:<br /><br />\[<br />-1 - (-6) = -1 + 6 = 5<br />\]<br /><br />So the expression becomes:<br /><br />\[<br />\text{Average value} = \frac{1}{5} \int_{-6}^{-1} \sin(x) \, dx<br />\]<br /><br />Next, we need to evaluate the integral \( \int_{-6}^{-1} \sin(x) \, dx \). We know that the integral of \( \sin(x) \) is \( -\cos(x) \):<br /><br />\[<br />\int \sin(x) \, dx = -\cos(x) + C<br />\]<br /><br />Applying the limits of integration from \(-6\) to \(-1\):<br /><br />\[<br />\int_{-6}^{-1} \sin(x) \, dx = \left[ -\cos(x) \right]_{-6}^{-1} = -\cos(-1) - (-\cos(-6))<br />\]<br /><br />Since \(\cos(-x) = \cos(x)\), this simplifies to:<br /><br />\[<br />-\cos(-1) + \cos(-6) = -\cos(1) + \cos(6)<br />\]<br /><br />Now, substitute this back into the expression for the average value:<br /><br />\[<br />\text{Average value} = \frac{1}{5} \left( -\cos(1) + \cos(6) \right)<br />\]<br /><br />Thus, the average value of \( \sin(x) \) on the interval \([-6, -1]\) is:<br /><br />\[<br />\boxed{\frac{1}{5} \left( \cos(6) - \cos(1) \right)}<br />\]