Free Math Worksheets

- First determine the width of each rectangle. - For an upper sum of an increasing function, decide whether to use the left or right endpoint of each subinterval. - Express each sample point in terms of the summation index. - Use the formula for the sum of the first \(n\) squares to simplify the larger sum.

1. For \(n=4\), the subinterval width is \(\Delta x=\frac{4-0}{4}=1\). Because \(f\) is increasing on \([0,4]\), each upper rectangle uses the right endpoint \(x_k=k\). Thus, \(U_4=\sum_{k=1}^{4} f(x_k)\Delta x=\sum_{k=1}^{4}\frac{1}{2}k^2\cdot 1=\frac{1}{2}(1^2+2^2+3^2+4^2)=15\). 2. For \(n=8\), \(\Delta x=\frac{4}{8}=0.5\), and the right endpoints are \(x_k=0.5k\). Therefore, \(U_8=\sum_{k=1}^{8}f(0.5k)\cdot 0.5=\sum_{k=1}^{8}\frac{1}{16}k^2\). Using \(\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}\), \(U_8=\frac{1}{16}\cdot\frac{8\cdot 9\cdot 17}{6}=\frac{51}{4}=12.75\).

1) \(U_4=\sum_{k=1}^{4}\frac{1}{2}k^2=15\) 2) \(U_8=\sum_{k=1}^{8}\frac{1}{16}k^2=12.75\)

- Determine whether the function is increasing or decreasing on the interval. This tells you which endpoint gives each rectangle's minimum height. - Write the subinterval width in terms of the number of rectangles. - Express the \(x\)-coordinate of the \(k\)th sample point. - Substitute that sample point into the function to get the height of the \(k\)th rectangle.

1. The subinterval width is \(\Delta x=\frac{2-0}{5}=0.4\). Because \(f\) is decreasing on \([0,2]\), the minimum on each subinterval occurs at its right endpoint, \(x_k=0.4k\). Therefore, \(L_5=\sum_{k=1}^{5}\left(4-(0.4k)^2\right)\cdot 0.4\). The function values at \(0.4\), \(0.8\), \(1.2\), \(1.6\), and \(2\) are \(3.84\), \(3.36\), \(2.56\), \(1.44\), and \(0\). Their sum is \(11.2\), so \(L_5=11.2\cdot 0.4=4.48\). 2. For \(n\) subintervals, \(\Delta x=\frac{2}{n}\), and the right endpoints are \(x_k=\frac{2k}{n}\). Thus, \(L_n=\sum_{k=1}^{n}f\left(\frac{2k}{n}\right)\frac{2}{n}=\sum_{k=1}^{n}\left(4-\left(\frac{2k}{n}\right)^2\right)\frac{2}{n}\).

1) \(L_5=4.48\) 2) \(L_n=\sum_{k=1}^{n}\left(4-\frac{4k^2}{n^2}\right)\frac{2}{n}\)

- Combine partial grid squares to estimate whole square units. - For a midpoint Riemann sum, use the midpoint of each equal subinterval as the sample point. - For the exact value, find an antiderivative and apply the Fundamental Theorem of Calculus.

1. For a), counting full and partial grid squares gives a reasonable estimate of about \(14.5\) to \(15\) square units. 2. For b), \(\Delta x=2\), and the midpoints are \(x=1\) and \(x=3\). Since \(f(1)=2.75\) and \(f(3)=4.75\), \(M_2=2(2.75+4.75)=15\). 3. For c), an antiderivative is \(F(x)=-\frac{1}{12}x^3+x^2+x\). Therefore, \(\int_{0}^{4}f(x)\,dx=F(4)-F(0)=-\frac{16}{3}+20=\frac{44}{3}\approx14.67\). 4. Both estimates are close to the exact value; the midpoint estimate is about \(0.33\) square unit too large.

a) About \(14.5\) to \(15\) square units b) \(15\) square units c) \(\frac{44}{3}\) square units, or approximately \(14.67\) square units

- Check the scale of each axis before reading coordinates. - Areas below the x-axis contribute negatively to a definite integral. - Estimate areas by counting grid squares or using simple geometric shapes. - A horizontal asymptote is a horizontal line approached by the graph toward the ends of the displayed behavior.

1. From the graph, the y-intercept is \((0, -1)\); the x-intercepts are approximately \((0.3, 0)\) and \((6.5, 0)\); the local maximum is approximately \((2, 1.9)\); the inflection point is approximately \((4, 1.2)\); and the horizontal asymptote is \(y=-1\). 2. Estimating signed areas from the grid gives \(\int_{-1}^{1}f(x)\,dx\approx-3.4\) and \(\int_{1}^{6}f(x)\,dx\approx6.4\). 3. Since \(|6.4|>|-3.4|\), the second integral has the greater absolute value.

a) y-intercept \((0, -1)\); x-intercepts approximately \((0.3, 0)\) and \((6.5, 0)\); local maximum approximately \((2, 1.9)\); inflection point approximately \((4, 1.2)\); horizontal asymptote \(y=-1\) b) \(\int_{-1}^{1}f(x)\,dx\approx-3.4\); \(\int_{1}^{6}f(x)\,dx\approx6.4\) c) \(\int_{1}^{6}f(x)\,dx\) has the greater absolute value.