Free Math Worksheets

- Use the average-value formula. - Set the function equal to the average value. - Solve the resulting equation. - Check that each solution lies in the given interval.

1. \(\int_0^3(x-1)^2\,\text{d}x=\left[\frac{(x-1)^3}{3}\right]_0^3=3\). 2. The average value is \(\frac{1}{3-0}(3)=1\). 3. Set \((x_0-1)^2=1\). Then \(x_0-1=\pm1\), so \(x_0=0\) or \(x_0=2\). 4. Both values lie in \([0,3]\).

\(x_0=0\) and \(x_0=2\)

- Use the average-value formula for a function. - Find an antiderivative of the sinusoidal model. - Interpret the integral unit before dividing by time. - A sinusoid averages to zero over one full period.

1. The average on \([10,22]\) is \(\frac{1}{12}\int_{10}^{22}T(t)\,\text{d}t\). 2. An antiderivative is \(-\frac{120}{\pi}\cos\left(\frac{\pi}{12}(t-10)\right)+22t\). 3. The integral equals \(\frac{240}{\pi}+264\), so the average is \(22+\frac{20}{\pi}\approx28.37\,{}^\circ\text{C}\). 4. The expression in part b is the average temperature over the full 24-hour day. The sine term averages to zero over one full period, so the value is \(22\,{}^\circ\text{C}\).

a) \(22+\frac{20}{\pi}\,{}^\circ\text{C}\approx28.37\,{}^\circ\text{C}\) b) The average temperature over the entire day; its value is \(22\,{}^\circ\text{C}\).

- Differentiate using the product rule. - Find the zeros of the profile to determine the full base interval. - Use the average-value formula. - Multiply the cross-sectional area by the perpendicular length.

1. By the product rule, \(F'(x)=(4x+20)e^{-0.2x}+(2x^2+20x+50)(-0.2)e^{-0.2x}\). 2. Simplifying gives \(F'(x)=(10-0.4x^2)e^{-0.2x}=f(x)\), so \(F\) is an antiderivative. 3. The zeros satisfy \(10-0.4x^2=0\), so the base extends from \(x=-5\) to \(x=5\). 4. The cross-sectional area is \(A=\int_{-5}^{5}f(x)\,\text{d}x=F(5)-F(-5)=\frac{200}{e}\approx73.58\,\text{m}^2\). 5. The average height is \(\frac{A}{10}=\frac{20}{e}\approx7.36\,\text{m}\). 6. The volume is \(50A=\frac{10{,}000}{e}\approx3678.79\,\text{m}^3\).

a) \(F'(x)=f(x)\) b) \(\frac{20}{e}\,\text{m}\approx7.36\,\text{m}\) c) \(\frac{10{,}000}{e}\,\text{m}^3\approx3678.79\,\text{m}^3\)

- Find the average value first. - Set the function equal to that value. - Use the symmetry of sine on \([0,\pi]\). - Use radian mode for numerical approximations.

1. \(\int_0^\pi(\sin x+1)\,\text{d}x=[-\cos x+x]_0^\pi=\pi+2\). 2. The average value is \(\frac{\pi+2}{\pi}=1+\frac{2}{\pi}\). 3. Set \(\sin c+1=1+\frac{2}{\pi}\), so \(\sin c=\frac{2}{\pi}\). 4. On \([0,\pi]\), the solutions are \(c=\arcsin\left(\frac{2}{\pi}\right)\) and \(c=\pi-\arcsin\left(\frac{2}{\pi}\right)\). 5. Numerically, these are approximately \(0.690\) and \(2.451\).

\(c=\arcsin\left(\frac{2}{\pi}\right)\approx0.690\) and \(c=\pi-\arcsin\left(\frac{2}{\pi}\right)\approx2.451\)

- Approximate the area under the data with trapezoids. - Divide the estimated integral by the total time. - Compute the ordinary mean separately. - Compare how the two methods weight the endpoint measurements.

1. With step size \(2\,\text{h}\), the trapezoidal estimate is \(\int_0^{10}c(t)\,\text{d}t\approx\frac{2}{2}(0+2(4.2)+2(6.8)+2(5.1)+2(3.4)+2.1)=41.1\,\text{mg}\cdot\text{h/L}\). 2. The estimated average concentration is \(\frac{41.1}{10}=4.11\,\text{mg/L}\). 3. The arithmetic mean is \(\frac{0+4.2+6.8+5.1+3.4+2.1}{6}=3.6\,\text{mg/L}\). 4. The trapezoidal average is higher because the endpoint values receive half the weight of the interior values in the time-weighted estimate.

Trapezoidal average: \(4.11\,\text{mg/L}\) Arithmetic mean: \(3.6\,\text{mg/L}\)

- Use the trapezoidal rule to estimate the integral. - Divide the estimated area by the interval length. - Compute the arithmetic mean independently. - Relate average value to an equal-area rectangle.

1. With step size \(3\,\text{h}\), \(\int_6^{18}I(t)\,\text{d}t\approx\frac{3}{2}(150+2(600)+2(950)+2(700)+200)=7275\,\text{W}\cdot\text{h/m}^2\). 2. The estimated average is \(\frac{7275}{12}=606.25\,\text{W/m}^2\). 3. The arithmetic mean is \(\frac{150+600+950+700+200}{5}=520\,\text{W/m}^2\). 4. Geometrically, \(606.25\) is the height of a rectangle over \([6,18]\) with the same area as the trapezoidal approximation under the irradiance curve.

1. \(606.25\,\text{W/m}^2\) 2. The arithmetic mean is \(520\,\text{W/m}^2\), which is lower. 3. The average is the height of an equal-area rectangle over the time interval.

- Use the maximum value of sine. - Energy is the integral of power over time. - Divide the total energy by the interval length to find average power.

1. The sine function reaches its maximum when \(\frac{\pi}{12}t=\frac{\pi}{2}\), so \(t=6\). The maximum is \(P(6)=800\,\text{W}\), at noon. 2. The energy is \(E=\int_0^{12}800\sin\left(\frac{\pi}{12}t\right)\,\text{d}t\). 3. \(E=\left[-\frac{9600}{\pi}\cos\left(\frac{\pi}{12}t\right)\right]_0^{12}=\frac{19{,}200}{\pi}\approx6111.55\,\text{Wh}\). 4. The average power is \(\frac{E}{12}=\frac{1600}{\pi}\approx509.30\,\text{W}\).

a) \(800\,\text{W}\) at noon b) \(\frac{19{,}200}{\pi}\,\text{Wh}\approx6111.55\,\text{Wh}\) c) \(\frac{1600}{\pi}\,\text{W}\approx509.30\,\text{W}\)

- Use the product rule and chain rule to differentiate \(f\). - A maximum can occur where the first derivative changes from positive to negative. - To verify an antiderivative, differentiate it. - Use the average-value formula for a function on an interval.

1. By the product and chain rules, \(f'(t)=(10-2t)e^{-0.2t}\). Since the exponential factor is positive, \(f'(t)=0\) when \(10-2t=0\), so \(t_{\max}=5\) hours. The derivative changes from positive to negative there. 2. \(f(5)=50e^{-1}\approx 18.39\,\text{mg/L}\). 3. Differentiate \(F\): \(F'(t)=-50\left[e^{-0.2t}-0.2(t+5)e^{-0.2t}\right]=10te^{-0.2t}=f(t)\). Therefore, \(F\) is an antiderivative of \(f\). 4. The average value on \([0,20]\) is \(\frac{1}{20}\int_0^{20}f(t)\,dt=\frac{F(20)-F(0)}{20}\). Since \(F(20)=-1250e^{-4}\approx-22.89\) and \(F(0)=-250\), the average is \(\frac{-22.89+250}{20}\approx 11.36\,\text{mg/L}\).

a) \(t_{\max}=5\) hours b) About \(18.39\,\text{mg/L}\) c) \(F'(t)=f(t)\), so \(F\) is an antiderivative of \(f\) d) About \(11.36\,\text{mg/L}\)