Free math worksheets

- Identify the width and height of the shaded rectangle. - Multiply rate by elapsed time. - Check how the time units cancel in the product.

1. The area under a constant rate graph is a rectangle. 2. Its width is \(4\,\text{s}\) and its height is \(3\,\frac{\text{L}}{\text{s}}\). 3. The accumulated amount is \(4\,\text{s}\cdot 3\,\frac{\text{L}}{\text{s}}=12\,\text{L}\).

\(12\,\text{L}\)

- Split the graph where the rate changes. - Find the rectangular area for each constant-rate interval. - Add the two accumulated amounts and keep the resulting quantity in gallons.

1. During the first \(2\) minutes, the rectangular area is \(2\,\text{min}\cdot 2\,\frac{\text{gal}}{\text{min}}=4\,\text{gal}\). 2. During the next \(1\) minute, the rectangular area is \(1\,\text{min}\cdot 5\,\frac{\text{gal}}{\text{min}}=5\,\text{gal}\). 3. Add the accumulated amounts: \(4\,\text{gal}+5\,\text{gal}=9\,\text{gal}\).

\(9\,\text{gal}\)

- Match each time interval with the speed at its left endpoint. - Every rectangle has the same width here. - Multiply each speed by its interval width, then add the resulting distances.

1. The three intervals are \([0,1]\), \([1,2]\), and \([2,3]\), each with width \(1\,\text{s}\). 2. The left-endpoint speeds are \(2\), \(3\), and \(5\,\frac{\text{m}}{\text{s}}\). 3. The estimate is \(1\cdot 2+1\cdot 3+1\cdot 5=10\,\text{m}\).

\(10\,\text{m}\)

- Read the height of each shaded rectangle from the graph. - Every rectangle has width \(1\) minute. - Think about whether a decreasing graph lies above or below a rectangle whose height is chosen at the right edge.

a) The right-endpoint rectangle heights are \(5\), \(4\), \(3\), and \(2\,\frac{\text{L}}{\text{min}}\). With width \(1\,\text{min}\), the estimate is \(5+4+3+2=14\,\text{L}\). b) Because the rate decreases throughout the interval, each right-endpoint height is below the graph over most of its rectangle, while each left-endpoint height is above the graph over most of its rectangle. Therefore the right-endpoint estimate is smaller.

a) \(14\,\text{L}\) b) For a decreasing rate, right-endpoint rectangles lie below the graph while left-endpoint rectangles lie above it, so the right-endpoint estimate is smaller.

- Find the width of each time interval before multiplying by its rate. - The intervals do not all have the same width. - Add the accumulated amounts from the three intervals.

1. On \([0,1]\), the accumulated amount is \(1\cdot 4=4\,\text{L}\). 2. On \([1,2.5]\), the interval width is \(1.5\) minutes, so the amount is \(1.5\cdot 2=3\,\text{L}\). 3. On \([2.5,4]\), the interval width is also \(1.5\) minutes, so the amount is \(1.5\cdot 5=7.5\,\text{L}\). 4. The total is \(4+3+7.5=14.5\,\text{L}\).

\(14.5\,\text{L}\)

- Split the region under the graph where its shape changes. - Identify the familiar geometric shape in each section. - Because velocity is nonnegative, each geometric area contributes positively to distance traveled.

1. From \(0\) to \(2\) seconds, the region is a triangle with area \(\frac{1}{2}\cdot 2\cdot 4=4\,\text{m}\). 2. From \(2\) to \(5\) seconds, the region is a rectangle with area \(3\cdot 4=12\,\text{m}\). 3. From \(5\) to \(6\) seconds, the region is a triangle with area \(\frac{1}{2}\cdot 1\cdot 4=2\,\text{m}\). 4. The total distance is \(4+12+2=18\,\text{m}\).

\(18\,\text{m}\)

- Identify the two parallel vertical heights of the trapezoid. - Use the horizontal time interval as the trapezoid’s width. - Check the units produced by rate multiplied by time.

1. The region under the rate graph is a trapezoid whose parallel sides have lengths \(2\) and \(6\), and whose width is \(4\) minutes. 2. Its area is \(\frac{2+6}{2}\cdot 4=16\). 3. Multiplying \(\frac{\text{L}}{\text{min}}\) by minutes gives liters, so the volume added is \(16\,\text{L}\).

\(16\,\text{L}\)

- Convert each constant-rate interval into a rectangular accumulated amount. - Express the unknown interval’s contribution in terms of \(r\). - The two interval contributions must add to the stated total.

1. The first interval contributes \(2\cdot 3=6\) units. 2. The second interval has width \(3\) hours, so it contributes \(3r\) units. 3. Use the total: \(6+3r=18\). 4. Then \(3r=12\), so \(r=4\,\frac{\text{units}}{\text{h}}\).

\(r=4\,\frac{\text{units}}{\text{h}}\)

- List the left endpoints for each partition before evaluating the rate. - Multiply each sampled rate by the common rectangle width. - For an increasing graph, compare each left-endpoint rectangle height with the curve over that interval.

a) With width \(1\), the left endpoints are \(0\) and \(1\). The estimate is \(1[r(0)+r(1)]=1(1+2)=3\). b) With width \(0.5\), the left endpoints are \(0\), \(0.5\), \(1\), and \(1.5\). The corresponding rates are \(1\), \(1.25\), \(2\), and \(3.25\). The estimate is \(0.5(1+1.25+2+3.25)=3.75\). c) The rate is increasing, so left-endpoint rectangles lie below the curve. Refining the partition replaces some coarse rectangle heights with larger later values, producing a larger lower estimate.

a) \(3\) b) \(3.75\) c) Because the rate is increasing, both are lower estimates, and the finer left-endpoint partition gives the larger one.

- Decide which endpoint gives the smaller height on each interval when the rate is increasing. - All three rectangles have width \(1\) minute. - Use one set of endpoint heights for the lower bound and the other set for the upper bound.

1. Because the rate is increasing, left-endpoint rectangles give a lower bound. Their heights are \(1\), \(2\), and \(4\), so the lower bound is \(1+2+4=7\) units. 2. Right-endpoint rectangles give an upper bound. Their heights are \(2\), \(4\), and \(7\), so the upper bound is \(2+4+7=13\) units. 3. Thus the accumulated amount lies between \(7\) and \(13\) units under the stated monotonicity assumption.

Lower bound: \(7\) units Upper bound: \(13\) units

- Equal starting and ending rates do not determine what happens between those times. - Break each region under the rate graph into familiar geometric shapes. - Compare the two total areas rather than only the endpoint heights.

1. For machine a), the rate is constantly \(2\) units per minute, so the rectangular area is \(4\cdot 2=8\) units. 2. For machine b), view the region as the same \(4\)-by-\(2\) rectangle plus a triangle above it with base \(4\) minutes and height \(2\) units per minute. 3. Machine b) accumulates \(8+\frac{1}{2}\cdot 4\cdot 2=12\) units. 4. Therefore, machine b) accumulates \(12-8=4\) more units, even though the two machines have the same initial and final rates.

Machine a) accumulates \(8\) units. Machine b) accumulates \(12\) units. Machine b) accumulates \(4\) more units.

- For each trapezoid, average the two endpoint heights and multiply by the interval width. - Keep the four-interval and two-interval calculations separate. - Compare which visible peaks are represented by each partition. - Multiply the power unit by the time unit to determine the accumulated-energy unit.

a) The four trapezoid areas are \(\frac{1+3}{2}(1)=2\), \(\frac{3+2}{2}(1)=2.5\), \(\frac{2+5}{2}(1)=3.5\), and \(\frac{5+3}{2}(1)=4\). Their total is \(12\,\text{kWh}\). b) Using only \(t=0,2,4\), the two trapezoids have width \(2\) hours. The estimate is \(2\cdot\frac{1+2}{2}+2\cdot\frac{2+3}{2}=8\,\text{kWh}\). c) The coarse estimate skips the higher intermediate power values at \(t=1\) and \(t=3\), so its straight segments cut below substantial parts of the displayed model. Power in kilowatts multiplied by time in hours gives kilowatt-hours.

a) \(12\,\text{kWh}\) b) \(8\,\text{kWh}\) c) The coarse estimate misses the intermediate peaks, so it is smaller. The accumulated energy unit is kilowatt-hours.