Free Math Worksheets

- Read the constant function value from the graph. - Find the width and height of the rectangle. - Use a negative signed area for a region below the \(x\)-axis.

1. The graph is the horizontal line \(y=-2\), so \(f(x)=-2\). 2. On \([-2,3]\), the region between the graph and the \(x\)-axis is a rectangle below the axis. 3. Its width is \(3-(-2)=5\), and its height is \(2\). 4. Because it lies below the \(x\)-axis, its signed contribution is \(-5\cdot 2=-10\).

\(\int_{-2}^{3}f(x)\,dx=-10\)

- Read the function value from the graph. - Find the width of the integration interval. - Interpret the region as a rectangle.

1. The graph is the horizontal line \(y=1.5\), so the function is constant and positive. 2. The interval width is \(1-(-3)=4\). 3. The region is a rectangle with width \(4\) and height \(1.5\), so the integral is \(4\cdot 1.5=6\).

\(\int_{-3}^{1}f(x)\,dx=6\)

- Read the left and right boundaries of the shaded region from the \(x\)-axis. - A definite integral represents net signed area between a graph and the \(x\)-axis. - Use the left boundary as the lower limit and the right boundary as the upper limit.

1. The function is \(f(x)=\frac{1}{2}x^2+1\). 2. The shaded region extends from \(x=0\) to \(x=2\). 3. Therefore, the net signed area is represented by \(\int_{0}^{2}\left(\frac{1}{2}x^2+1\right)\,dx\).

\(\int_{0}^{2}\left(\frac{1}{2}x^2+1\right)\,dx\)

- Net signed area uses the signs of the function values. - One definite integral over the entire interval automatically combines regions above and below the \(x\)-axis.

1. The function is \(f(x)=\frac{1}{2}x-1\). 2. The interval extends from \(x=0\) to \(x=4\). 3. Therefore, the net signed area is represented by \(\int_{0}^{4}\left(\frac{1}{2}x-1\right)\,dx\). The function changes sign at \(x=2\), and the integral automatically combines the negative and positive contributions.

\(\int_{0}^{4}\left(\frac{1}{2}x-1\right)\,dx\)

- Sketch the graph on the given interval. - Identify the geometric figure formed by the graph and the \(x\)-axis. - Recall the area formulas for rectangles, triangles, and trapezoids. - Relate the endpoint function values to the side lengths of the figure.

1. The graph of \(f(x)=0.5x+3\) is a line. On \([-1,4]\), its endpoint values are \(f(-1)=2.5\) and \(f(4)=5\), so the region between the graph and the \(x\)-axis is a trapezoid entirely above the axis. 2. The parallel sides of the trapezoid have lengths \(2.5\) and \(5\). The distance between them is the interval length, \(4-(-1)=5\). 3. Its area is \(A=\frac{2.5+5}{2}\cdot 5=3.75\cdot 5=18.75\). 4. Because the region lies above the \(x\)-axis, the definite integral equals this area. Therefore, the integral is \(18.75\).

\(\int_{-1}^{4} (0.5x + 3)\,dx=18.75\)

- The variable after \(d\) is the variable of integration. - Treat all other letters as constants. - Use the power rule in the selected variable. - Integrating a constant over an interval multiplies it by the interval length.

1. For a), integrate with respect to \(x\): \(\int(4x^3y-2y^2)\,dx=x^4y-2xy^2\). Evaluating from \(1\) to \(3\) gives \(80y-4y^2\). 2. For b), integrate with respect to \(y\): \(\int(4x^3y-2y^2)\,dy=2x^3y^2-\frac{2}{3}y^3\). Evaluating from \(1\) to \(3\) gives \(16x^3-\frac{52}{3}\). 3. For c), \(a^2\) is constant with respect to \(x\): \(\int_0^1a^2\,dx=a^2\).

a) \(80y-4y^2\) b) \(16x^3-\frac{52}{3}\) c) \(a^2\)

- Identify the variable after \(d\). - Treat every other letter as a constant. - After applying the bounds, the result should not contain the variable of integration. - Distinguish \(e^z w\) from \(e^{zw}\).

1. For a), integrate with respect to \(t\): \(\left[\frac{t^4}{4}\right]_0^k=\frac{k^4}{4}\). 2. For b), integrate with respect to \(k\): \(\left[\frac{k^4}{4}\right]_0^t=\frac{t^4}{4}\). 3. For c), \(w\) is constant with respect to \(z\): \(w[e^z]_0^1=w(e-1)\). 4. For d), \(e^z\) is constant with respect to \(w\): \(e^z\left[\frac{w^2}{2}\right]_0^1=\frac{e^z}{2}\).

a) \(\frac{k^4}{4}\) b) \(\frac{t^4}{4}\) c) \(w(e-1)\) d) \(\frac{e^z}{2}\)

- Interpret the definite integral as net signed area. - Identify which regions lie above and below the \(x\)-axis. - Divide the regions into triangles or other simple shapes. - Find the \(x\)-intercepts before combining the signed areas.

1. Split the interval at the zeros \(x=-2\) and \(x=3\). 2. On \([-4,-2]\), the graph forms a triangle below the \(x\)-axis with base \(2\) and height \(2\). Its signed contribution is \(-\frac{1}{2}\cdot 2\cdot 2=-2\). 3. On \([-2,3]\), the graph forms a triangle above the \(x\)-axis with base \(5\) and height \(2\). Its contribution is \(\frac{1}{2}\cdot 5\cdot 2=5\). 4. On \([3,4]\), the graph forms a triangle below the \(x\)-axis with base \(1\) and height \(2\). Its signed contribution is \(-\frac{1}{2}\cdot 1\cdot 2=-1\). 5. Therefore, \(\int_{-4}^{4}f(x)\,dx=-2+5-1=2\).

\(2\)

- Interpret the definite integral as net signed area. - Separate regions above and below the \(x\)-axis. - Divide the regions into triangles and rectangles. - Combine the areas with the correct signs.

1. Divide the regions between the graph and the \(x\)-axis into triangles and a rectangle. 2. Above the \(x\)-axis, the triangle on \([-3,0]\) has area \(\frac{1}{2}\cdot 3\cdot 2=3\), and the triangle on \([4,5]\) has area \(\frac{1}{2}\cdot 1\cdot 2=1\). The total positive contribution is \(4\). 3. Below the \(x\)-axis, the triangle on \([0,1]\) has area \(1\), the rectangle on \([1,3]\) has area \(4\), and the triangle on \([3,4]\) has area \(1\). The total magnitude of the negative contribution is \(6\). 4. Therefore, \(\int_{-3}^{5}f(x)\,dx=4-6=-2\).

\(-2\)

- Compare the horizontal boundaries in each figure with the limits of integration. - Identify whether each graph is linear, quadratic, or reciprocal. - Check key features such as an intercept, a vertex, or points on the graph. - Distinguish a shifted parabola from \(y=x^2\).

1. Figure a) shows a line with \(y\)-intercept \(1\), slope \(0.5\), and boundaries at \(x=0\) and \(x=4\). It matches integral (1). 2. Figure b) shows an upward-opening parabola with vertex \((2,0)\), so its equation is \(y=(x-2)^2\). The boundaries are \(x=0\) and \(x=2\), so it matches integral (2). 3. Figure c) shows a reciprocal graph through \((1,4)\) and \((4,1)\). For \(y=\frac{k}{x}\), the point \((1,4)\) gives \(k=4\). The boundaries are \(x=1\) and \(x=4\), so it matches integral (3).

a) Integral (1) b) Integral (2) c) Integral (3)

- Identify the regions above and below the \(x\)-axis. - Find the dimensions of the two triangles. - Add their signed areas.

1. The graph crosses the \(x\)-axis at \(x=2\). 2. On \([-2,2]\), the graph forms a triangle above the axis with base \(4\) and height \(2\). Its contribution is \(\frac{1}{2}\cdot 4\cdot 2=4\). 3. On \([2,4]\), the graph forms a triangle below the axis with base \(2\) and height \(1\). Its signed contribution is \(-\frac{1}{2}\cdot 2\cdot 1=-1\). 4. The integral is \(4-1=3\).

\(\int_{-2}^{4}f(x)\,dx=3\)

- Divide the region into triangles. - Find where the graph changes sign. - Add the signed areas above and below the \(x\)-axis.

1. The graph has zeros at \(x=-3\) and \(x=1\), with a maximum at \((-1,2)\). 2. On \([-3,1]\), the graph forms a triangle above the \(x\)-axis with base \(4\) and height \(2\). Its area is \(\frac{1}{2}\cdot 4\cdot 2=4\). 3. On \([1,2]\), the graph forms a triangle below the axis with base \(1\) and height \(1\). Its signed contribution is \(-\frac{1}{2}\cdot 1\cdot 1=-0.5\). 4. Therefore, \(\int_{-3}^{2}f(x)\,dx=4-0.5=3.5\).

\(\int_{-3}^{2}f(x)\,dx=3.5\)

- Interpret the integral as net signed area. - Find the base and height of each triangle. - Use a negative contribution for the region below the \(x\)-axis.

1. The graph crosses the \(x\)-axis at \(x=1\). 2. On \([-2,1]\), the graph forms a triangle above the axis with base \(3\) and height \(3\). Its contribution is \(\frac{1}{2}\cdot 3\cdot 3=4.5\). 3. On \([1,3]\), the graph forms a triangle below the axis with base \(2\) and height \(2\). Its signed contribution is \(-\frac{1}{2}\cdot 2\cdot 2=-2\). 4. Therefore, \(\int_{-2}^{3}f(x)\,dx=4.5-2=2.5\).

\(\int_{-2}^{3}f(x)\,dx=2.5\)

- Find where the graph crosses the \(x\)-axis. - Divide the interval into regions that lie entirely above or below the axis. - Use a negative signed area for the region below the axis.

1. The graph crosses the \(x\)-axis at \(x=1\). It lies below the axis on \([-1,1]\) and above it on \([1,2]\). 2. The triangle below the axis has base \(2\) and height \(2\), so its signed contribution is \(-\frac{1}{2}\cdot 2\cdot 2=-2\). 3. The triangle above the axis has base \(1\) and height \(1\), so its contribution is \(\frac{1}{2}\cdot 1\cdot 1=0.5\). 4. Therefore, \(\int_{-1}^{2}f(x)\,dx=-2+0.5=-1.5\).

The integral is \(-1.5\).

- Use the graph's symmetry to reduce the work. - Identify the positive and negative triangular regions. - Check whether their signed contributions cancel.

1. The graph is symmetric about the \(y\)-axis and crosses the \(x\)-axis at \(x=-1\) and \(x=1\). 2. By symmetry, calculate the integral on \([0,2]\) and double it. 3. On \([0,1]\), the graph forms a triangle below the axis with base \(1\) and height \(1\), contributing \(-0.5\). 4. On \([1,2]\), the graph forms a congruent triangle above the axis, contributing \(0.5\). 5. Thus, \(\int_{0}^{2}f(x)\,dx=-0.5+0.5=0\), and \(\int_{-2}^{2}f(x)\,dx=2\cdot 0=0\).

The integral is \(0\).

- Find the zeros of the function and determine where its graph is above or below the \(x\)-axis. - Remember that regions below the \(x\)-axis contribute negatively to a definite integral. - Compare the positive and negative signed-area contributions before calculating.

1. On \([0,2]\), the graph lies on or below the \(x\)-axis because \(x^2-4\leq 0\). It is below the axis except at \(x=2\), so the definite integral is negative. 2. On \([0,4]\), the graph is below the \(x\)-axis from \(0\) to \(2\) and above it from \(2\) to \(4\). The positive signed-area contribution is larger, so the integral should be positive. 3. An antiderivative is \(F(x)=\frac{1}{3}x^3-4x\). Therefore, \(\int_{0}^{2}(x^2-4)\,dx=F(2)-F(0)=\frac{8}{3}-8=-\frac{16}{3}\), and \(\int_{0}^{4}(x^2-4)\,dx=F(4)-F(0)=\frac{64}{3}-16=\frac{16}{3}\). Both calculations confirm the predictions.

1) \(\int_{0}^{2}f(x)\,dx\) is negative. 2) The graph’s key features indicate that \(\int_{0}^{4}f(x)\,dx\) is positive. 3) \(\int_{0}^{2}f(x)\,dx=-\frac{16}{3}\) and \(\int_{0}^{4}f(x)\,dx=\frac{16}{3}\).

- Split the integral where the function rule changes. - Check whether the graph crosses the \(x\)-axis within either piece. - Identify the rectangles and triangles formed by the graph. - Use a negative contribution for any region below the \(x\)-axis.

1. Split the interval at \(x=1\), where the formula changes, and at \(x=3\), where \(3-x=0\). 2. On \([-2,1]\), the graph forms a rectangle of width \(3\) and height \(2\), contributing \(3\cdot 2=6\). 3. On \([1,3]\), the graph forms a triangle above the \(x\)-axis with base \(2\) and height \(2\), contributing \(\frac{1}{2}\cdot 2\cdot 2=2\). 4. On \([3,5]\), the graph forms a congruent triangle below the \(x\)-axis, contributing \(-\frac{1}{2}\cdot 2\cdot 2=-2\). 5. The net signed area is \(6+2-2=6\).

\(6\)

- Use the listed points to analyze the connected line segments. - Divide the region into rectangles and triangles. - Find where the graph crosses the \(x\)-axis. - Treat areas below the \(x\)-axis as negative contributions.

1. The segment from \(B(2,2)\) to \(C(5,-1)\) has slope \(-1\), so its equation is \(y=-x+4\). It crosses the \(x\)-axis at \(x=4\). 2. Above the \(x\)-axis, the rectangle from \(x=0\) to \(x=2\) contributes \(2\cdot 2=4\), and the triangle from \(x=2\) to \(x=4\) contributes \(\frac{1}{2}\cdot 2\cdot 2=2\). 3. Below the \(x\)-axis, the triangle from \(x=4\) to \(x=5\) has area \(\frac{1}{2}\cdot 1\cdot 1=0.5\), and the rectangle from \(x=5\) to \(x=7\) has area \(2\cdot 1=2\). These contribute negatively. 4. The net signed area is \(4+2-0.5-2=3.5\).

\(\int_{0}^{7}f(x)\,dx=3.5\)

- Relate each definite integral to signed area between the graph and the \(x\)-axis. - Identify where the graph is above and below the axis. - Bound curved regions with triangles or rectangles. - Interpret the integral of the constant function \(1\) geometrically.

1. For a), the positive region on \([-2,0]\) is smaller than the negative region on \([0,3]\). The net signed area is negative, so the statement is true. Numerically, the integral is about \(-5.21\). 2. For b), the region below the \(x\)-axis contains the triangle with vertices \((0,0)\), \((2,-4)\), and \((3,0)\), whose area is \(\frac{1}{2}\cdot 3\cdot 4=6\). The curve lies below the two slanted sides of this triangle, so the magnitude of the negative area is greater than \(6\). Thus, \(\int_{0}^{3}f(x)\,dx<-6\), and the statement is false. 3. For c), \(\int_{-2}^{0}1\,dx=2\). The region under \(f\) contains the triangle with vertices \((-2,0)\), \((-1,2)\), and \((0,0)\), which has area \(2\). Since the graph lies above the triangle's slanted sides, \(\int_{-2}^{0}f(x)\,dx>2\). The statement is true.

a) True. The negative signed area on \([0,3]\) has greater magnitude than the positive signed area on \([-2,0]\). b) False. In fact, \(\int_{0}^{3}f(x)\,dx<-6\). c) True. \(\int_{-2}^{0}f(x)\,dx>2=\int_{-2}^{0}1\,dx\).

- Find the width, height, and area of the red rectangle. - Interpret the definite integral as area under the graph on this interval. - Use the grid to make lower rectangular estimates on \([-2,2]\) and on the two edge intervals.

1. The red rectangle has width \(6\), from \(x=-3\) to \(x=3\), and height \(1\). Its area is \(6\cdot 1=6\). 2. The graph is above the \(x\)-axis throughout \([-3,3]\), so the definite integral is the area under the graph. 3. On \([-2,2]\), the graph has height at least \(f(2)=f(-2)=1.2\). Therefore, the area under the graph on this central interval is greater than \(4\cdot 1.2=4.8\). 4. On the two edge intervals \([-3,-2]\) and \([2,3]\), the graph has height at least \(f(3)=f(-3)=0.7\). Their combined area is greater than \(2\cdot 0.7=1.4\). 5. Thus, the area under the graph is greater than \(4.8+1.4=6.2\), so \(\int_{-3}^{3}f(x)\,dx>6\).

The red rectangle has area \(6\). The area under the graph is greater than \(4\cdot 1.2+2\cdot 0.7=6.2\), using lower rectangular estimates from the grid. Therefore, \(\int_{-3}^{3}f(x)\,dx>6\).