Free Math Worksheets

- Use integration by parts for the product of a linear function and an exponential function. - Differentiate the linear factor and integrate the exponential factor. - Account for the inner derivative in \(e^{3x}\). - Factor out the exponential term when simplifying.

1. Use integration by parts with \(u=x+1\) and \(dv=e^{3x}\,dx\). Then \(du=dx\) and \(v=\frac{1}{3}e^{3x}\). 2. Therefore, \(\int(x+1)e^{3x}\,dx=\frac{1}{3}(x+1)e^{3x}-\frac{1}{3}\int e^{3x}\,dx\). 3. Evaluate the remaining integral: \(\frac{1}{3}(x+1)e^{3x}-\frac{1}{9}e^{3x}\). 4. Simplify to obtain \(F(x)=\left(\frac{x}{3}+\frac{2}{9}\right)e^{3x}\).

\(F(x)=\left(\frac{x}{3}+\frac{2}{9}\right)e^{3x}\)

- Choose the factor that becomes simpler when differentiated. - Account for the inner derivative when integrating \(\cos(2x)\). - Track the sign when integrating sine. - Differentiate your result to verify it.

1. Use integration by parts with \(u=4x\) and \(dv=\cos(2x)\,dx\). Then \(du=4\,dx\) and \(v=\frac{1}{2}\sin(2x)\). 2. Apply the formula: \(\int 4x\cos(2x)\,dx=2x\sin(2x)-2\int\sin(2x)\,dx\). 3. Since \(\int 2\sin(2x)\,dx=-\cos(2x)\), the result is \(F(x)=2x\sin(2x)+\cos(2x)\).

\(F(x)=2x\sin(2x)+\cos(2x)\)

- Differentiate the linear factor. - Use integration by parts. - Account for the inner factor when integrating \(\sin(2x)\) and \(\cos(2x)\). - Track the signs carefully.

1. Use integration by parts with \(u=2-x\) and \(dv=\sin(2x)\,dx\). Then \(du=-dx\) and \(v=-\frac{1}{2}\cos(2x)\). 2. Therefore, \(F(x)=-\frac{1}{2}(2-x)\cos(2x)-\frac{1}{2}\int\cos(2x)\,dx\). 3. Since \(\int\cos(2x)\,dx=\frac{1}{2}\sin(2x)\), \(F(x)=\left(\frac{x}{2}-1\right)\cos(2x)-\frac{1}{4}\sin(2x)\).

\(F(x)=\left(\frac{x}{2}-1\right)\cos(2x)-\frac{1}{4}\sin(2x)\)

- Choose the factor that becomes simpler when differentiated. - Track the negative sign when integrating sine. - Recall the sine and cosine values at \(0\) and \(\frac{\pi}{2}\). - Evaluate the complete integration-by-parts expression at both limits.

1. Choose \(u=2x-1\) and \(dv=\sin(x)\,dx\). Then \(du=2\,dx\) and \(v=-\cos(x)\). 2. Apply integration by parts: \(\int_0^{\frac{\pi}{2}}(2x-1)\sin(x)\,dx=[-(2x-1)\cos(x)]_0^{\frac{\pi}{2}}+2\int_0^{\frac{\pi}{2}}\cos(x)\,dx\). 3. Combine the terms: \([-(2x-1)\cos(x)+2\sin(x)]_0^{\frac{\pi}{2}}\). 4. Substitute the limits. The upper value is \(2\), and the lower value is \(1\), so the integral equals \(1\).

\(1\)

- Choose one exponent whose exponential factor remains simple under differentiation. - For substitution, look for an exponent whose derivative is a constant multiple of \(x\). - Integration by parts is useful when differentiating one factor simplifies it. - Check that each chosen exponent leads to the method requested.

1. Choose the exponent \(x\). Then \(\int_0^1xe^x\,dx\) is evaluated by integration by parts. With \(u=x\) and \(dv=e^x\,dx\), an antiderivative is \((x-1)e^x\). Thus \([(x-1)e^x]_0^1=1\). 2. Choose the exponent \(x^2\). Then \(\int_0^1xe^{x^2}\,dx\) is evaluated by substitution. Let \(u=x^2\), so \(du=2x\,dx\). Therefore, \(\int_0^1xe^{x^2}\,dx=\frac{1}{2}\int_0^1e^u\,du=\frac{e-1}{2}\).

One valid pair is: Exponent \(x\): integration by parts gives \(1\). Exponent \(x^2\): substitution gives \(\frac{e-1}{2}\).

- Write the integrand as a product with \(1\). - Differentiate the logarithmic factor and integrate the constant factor. - Simplify the remaining integral after applying integration by parts. - Use \(\ln(1)=0\) at the lower limit.

1. Use integration by parts with \(u=\ln(5t)\) and \(dv=dt\). Then \(du=\frac{1}{t}\,dt\) and \(v=t\). 2. Thus \(\int\ln(5t)\,dt=t\ln(5t)-\int 1\,dt=t\ln(5t)-t\). 3. Evaluate the accumulation function: \(I_{0.2}(x)=[t\ln(5t)-t]_{0.2}^{x}\). 4. Since \(\ln(5\cdot 0.2)=\ln(1)=0\), \(I_{0.2}(x)=x\ln(5x)-x+0.2\).

\(I_{0.2}(x)=x\ln(5x)-x+0.2\)

- Split the integrand into two factors that are both related to \(\ln(x)\). - Choose a factor whose antiderivative matches the other factor. - Identify the repeated integral after applying integration by parts. - Treat the integral as an unknown in an algebraic equation.

1. Let \(I=\int_1^e\frac{\ln(x)}{x}\,dx\). Choose \(u=\ln(x)\) and \(dv=\frac{1}{x}\,dx\). Then \(du=\frac{1}{x}\,dx\) and \(v=\ln(x)\). 2. Integration by parts gives \(I=[\ln^2(x)]_1^e-I\). 3. Therefore, \(2I=[\ln^2(x)]_1^e=\ln^2(e)-\ln^2(1)=1\). 4. Hence \(I=\frac{1}{2}\).

\(\frac{1}{2}\)

- Differentiate the polynomial factor so its degree decreases. - Use integration by parts on the remaining product again. - Account for the factor \(2\) when integrating \(e^{2x}\). - Evaluate the complete antiderivative at the limits.

1. Choose \(u=x^2\) and \(dv=e^{2x}\,dx\). Then \(du=2x\,dx\) and \(v=\frac{1}{2}e^{2x}\), so \(\int_0^1x^2e^{2x}\,dx=[\frac{1}{2}x^2e^{2x}]_0^1-\int_0^1xe^{2x}\,dx\). 2. For the remaining integral, use \(u=x\) and \(dv=e^{2x}\,dx\): \(\int xe^{2x}\,dx=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}\). 3. An antiderivative of the original integrand is \(F(x)=\left(\frac{1}{2}x^2-\frac{1}{2}x+\frac{1}{4}\right)e^{2x}\). 4. Evaluate: \(F(1)-F(0)=\frac{1}{4}e^2-\frac{1}{4}=\frac{e^2-1}{4}\).

\(\frac{e^2-1}{4}\)

- Differentiate the polynomial factor to lower its degree. - Apply integration by parts again to the remaining product. - Track the signs when integrating sine. - Use the sine and cosine values at \(0\) and \(\frac{\pi}{2}\).

1. Choose \(u=x^2\) and \(dv=\cos(x)\,dx\). Then \(du=2x\,dx\) and \(v=\sin(x)\): \(\int_0^{\frac{\pi}{2}}x^2\cos(x)\,dx=[x^2\sin(x)]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}2x\sin(x)\,dx\). 2. For the remaining integral, choose \(u=2x\) and \(dv=\sin(x)\,dx\). Then \(du=2\,dx\) and \(v=-\cos(x)\), so \(\int 2x\sin(x)\,dx=-2x\cos(x)+2\sin(x)\). 3. An antiderivative of the original integrand is \(F(x)=x^2\sin(x)+2x\cos(x)-2\sin(x)\). 4. Evaluate: \(F(\frac{\pi}{2})-F(0)=\frac{\pi^2}{4}-2\).

\(\frac{\pi^2}{4}-2\)

- Choose the factor that becomes simpler when differentiated. - Use the integration-by-parts formula. - Recall the values of \(\ln(e)\) and \(\ln(1)\). - Combine like terms at the end.

1. Let \(u=\ln(x)\) and \(dv=x^2\,dx\). Then \(du=\frac{1}{x}\,dx\) and \(v=\frac{x^3}{3}\). 2. Integration by parts gives \(\int x^2\ln(x)\,dx=\frac{x^3}{3}\ln(x)-\frac{1}{3}\int x^2\,dx\) \(=\frac{x^3}{3}\ln(x)-\frac{x^3}{9}\). 3. Evaluate from \(1\) to \(e\): \(\left[\frac{x^3}{3}\ln(x)-\frac{x^3}{9}\right]_1^e=\left(\frac{e^3}{3}-\frac{e^3}{9}\right)-\left(0-\frac{1}{9}\right)\) \(=\frac{2e^3+1}{9}\).

\(\frac{2e^3+1}{9}\)

- Differentiate the linear factor. - Be careful when finding an antiderivative of \(e^{-x}\). - Apply the bounds to each term carefully. - Simplify the exponential values at the end.

1. Let \(u=2x+1\) and \(dv=e^{-x}\,dx\). Then \(du=2\,dx\) and \(v=-e^{-x}\). 2. Apply integration by parts: \(\int_0^1(2x+1)e^{-x}\,dx=[-(2x+1)e^{-x}]_0^1+2\int_0^1e^{-x}\,dx\). 3. Evaluate: \([-(2x+1)e^{-x}]_0^1+[-2e^{-x}]_0^1\) \(=\left(-\frac{3}{e}+1\right)+\left(-\frac{2}{e}+2\right)=3-\frac{5}{e}\).

\(3-\frac{5}{e}\approx 1.1606\)

- Differentiate the factor that becomes simpler. - Compare what differentiation and integration do to a polynomial degree. - The goal is to make the remaining integral easier than the original.

1. For a), if \(u=e^{0.5x}\) and \(dv=(2x+4)\,dx\), then \(du=0.5e^{0.5x}\,dx\) and \(v=x^2+4x\). The new integral contains the quadratic factor \(x^2+4x\), so the polynomial degree increases from \(1\) to \(2\). 2. For b), choose \(u=2x+4\) and \(dv=e^{0.5x}\,dx\). Then \(du=2\,dx\) and \(v=2e^{0.5x}\). 3. Apply integration by parts: \(I=[(2x+4)2e^{0.5x}]_0^1-4\int_0^1e^{0.5x}\,dx\). 4. Evaluate: \(I=[4(x+2)e^{0.5x}]_0^1-[8e^{0.5x}]_0^1\) \(=(12\sqrt e-8)-(8\sqrt e-8)=4\sqrt e\).

a) The polynomial degree increases from \(1\) to \(2\). b) \(I=4\sqrt e\)

- Differentiate the logarithmic factor. - Apply integration by parts. - Use \(\ln(e)=1\) and \(\ln(1)=0\).

1. Let \(u=\ln(x)\) and \(dv=x\,dx\). Then \(du=\frac{1}{x}\,dx\) and \(v=\frac{x^2}{2}\). 2. Integration by parts gives \(F(x)=\frac{x^2}{2}\ln(x)-\frac{1}{2}\int x\,dx\) \(=\frac{x^2}{2}\ln(x)-\frac{x^2}{4}\). 3. Evaluate: \(F(e)-F(1)=\frac{e^2}{4}-\left(-\frac{1}{4}\right)=\frac{e^2+1}{4}\).

One antiderivative is \(F(x)=\frac{x^2}{2}\ln(x)-\frac{x^2}{4}\). The integral equals \(\frac{e^2+1}{4}\).

- Compare two members of the family to find parameter-independent points. - Determine whether the function is above or below the x-axis on \([0, 2]\). - Use integration by parts to find an antiderivative of the polynomial-exponential product. - Take the absolute value of the definite integral when finding area.

1. If two distinct parameter values \(a\) and \(b\) give the same function value, then \((x^2-ax)e^x=(x^2-bx)e^x\). Since \(e^x\ne0\), this reduces to \((a-b)x=0\). For \(a\ne b\), this requires \(x=0\), and \(f_a(0)=0\). Thus, the only common point is \((0, 0)\). 2. For \(a=2\), \(f_2(x)=x(x-2)e^x\le0\) on \([0, 2]\). Two applications of integration by parts give \(F(x)=(x^2-4x+4)e^x\) as an antiderivative. Hence, \(\int_0^2f_2(x)\,\text{d}x=F(2)-F(0)=-4\). The area is the absolute value, so it equals \(4\) square units.

1) \((0, 0)\) 2) \(4\) square units

- Apply integration by parts more than once. - Differentiating a power of \(\ln(t)\) lowers that power. - Track the signs when substituting the second integration-by-parts result. - Evaluate all logarithmic terms at the lower limit \(1\).

1. Use integration by parts with \(u=[\ln(t)]^2\) and \(dv=dt\). Then \(du=\frac{2\ln(t)}{t}\,dt\) and \(v=t\). 2. Therefore, \(\int[\ln(t)]^2\,dt=t[\ln(t)]^2-2\int\ln(t)\,dt\). 3. Apply integration by parts again to \(\int\ln(t)\,dt\), obtaining \(t\ln(t)-t\). 4. An antiderivative is \(t[\ln(t)]^2-2t\ln(t)+2t\). 5. Evaluate from \(1\) to \(x\). Since \(\ln(1)=0\), \(I_1(x)=x[\ln(x)]^2-2x\ln(x)+2x-2\).

\(I_1(x)=x[\ln(x)]^2-2x\ln(x)+2x-2\)

- Apply integration by parts more than once. - Track the signs when differentiating cosine. - When the original integral returns, move it to the same side of the equation. - Use the sine and cosine values at \(0\) and \(\pi\).

1. Let \(I=\int_0^\pi e^x\sin(x)\,dx\). Use integration by parts with \(u=\sin(x)\) and \(dv=e^x\,dx\): \(I=[e^x\sin(x)]_0^\pi-\int_0^\pi e^x\cos(x)\,dx\). The boundary term is \(0\), so \(I=-\int_0^\pi e^x\cos(x)\,dx\). 2. Apply integration by parts to the remaining integral with \(u=\cos(x)\) and \(dv=e^x\,dx\): \(\int_0^\pi e^x\cos(x)\,dx=[e^x\cos(x)]_0^\pi+I\). 3. Substitute into the first equation: \(I=-[e^x\cos(x)]_0^\pi-I\). Thus \(2I=-[e^x\cos(x)]_0^\pi\). 4. Evaluate the boundary term: \(-[e^\pi\cos(\pi)-e^0\cos(0)]=e^\pi+1\). Therefore, \(I=\frac{e^\pi+1}{2}\).

\(\frac{e^\pi+1}{2}\)

- Insert a factor of \(1\) to create a product. - Apply integration by parts again to the remaining logarithmic integral. - Track the minus sign carefully. - Recall the derivative of \(\ln(x)\).

1. Let \(u=(\ln(x))^2\) and \(dv=dx\). Then \(du=\frac{2\ln(x)}{x}\,dx\) and \(v=x\). 2. Thus, \(\int(\ln(x))^2\,dx=x(\ln(x))^2-2\int\ln(x)\,dx\). 3. Apply integration by parts again to \(\int\ln(x)\,dx\): \(\int\ln(x)\,dx=x\ln(x)-x\). 4. Therefore, an antiderivative is \(F(x)=x(\ln(x))^2-2x\ln(x)+2x\). 5. Evaluate: \([F(x)]_1^e=e-2\).

\(e-2\)

- Differentiate the polynomial factor so its degree decreases. - Apply integration by parts twice. - Track the signs when integrating sine and cosine. - Use the exact trigonometric values at \(0\) and \(\pi\).

1. Let \(u=x^2-2x\) and \(dv=\sin(x)\,dx\). Then \(du=(2x-2)\,dx\) and \(v=-\cos(x)\). 2. Therefore, \(\int(x^2-2x)\sin(x)\,dx=-(x^2-2x)\cos(x)+\int(2x-2)\cos(x)\,dx\). 3. Apply integration by parts again with \(u=2x-2\) and \(dv=\cos(x)\,dx\): \(\int(2x-2)\cos(x)\,dx=(2x-2)\sin(x)+2\cos(x)\). 4. An antiderivative is \(F(x)=-(x^2-2x)\cos(x)+(2x-2)\sin(x)+2\cos(x)\). 5. Evaluate: \(F(\pi)=\pi^2-2\pi-2\) and \(F(0)=2\), so the integral is \(\pi^2-2\pi-4\).

\(\pi^2-2\pi-4\approx -0.414\)

- Differentiate the exponential factor and integrate the trigonometric factor twice. - Treat the recurring integral as an unknown quantity. - Solve the resulting equation for that integral. - Use the exact sine and cosine values at the bounds.

1. Let \(I=\int e^{2x}\cos(x)\,dx\). With \(u=e^{2x}\) and \(dv=\cos(x)\,dx\), \(I=e^{2x}\sin(x)-2\int e^{2x}\sin(x)\,dx\). 2. Apply integration by parts to the remaining integral: \(\int e^{2x}\sin(x)\,dx=-e^{2x}\cos(x)+2I\). 3. Substitute this expression: \(I=e^{2x}\sin(x)+2e^{2x}\cos(x)-4I\). Thus, \(I=\frac{1}{5}e^{2x}(\sin(x)+2\cos(x))\). 4. Evaluate from \(0\) to \(\frac{\pi}{2}\): \(\frac{1}{5}e^\pi-\frac{2}{5}=\frac{e^\pi-2}{5}\).

\(\frac{e^\pi-2}{5}\approx 4.228\)

- View \(\cos^2(x)\) as a product of two cosine factors. - Use \(\sin^2(x)+\cos^2(x)=1\). - Solve algebraically when the original integral appears on both sides. - Use the endpoint values of sine and cosine.

1. Using integration by parts with \(u=\cos(x)\) and \(dv=\cos(x)\,dx\), let \(I=\int\cos^2(x)\,dx\). Then \(I=\sin(x)\cos(x)+\int\sin^2(x)\,dx\). 2. Since \(\sin^2(x)=1-\cos^2(x)\), \(I=\sin(x)\cos(x)+x-I\). Therefore, \(I=\frac{1}{2}(x+\sin(x)\cos(x))\). 3. Evaluate from \(0\) to \(\pi\): \(\left[\frac{1}{2}(x+\sin(x)\cos(x))\right]_0^\pi=\frac{\pi}{2}\).

\(\frac{\pi}{2}\)

- Find where the natural logarithm crosses the \(x\)-axis. - Use integration by parts to find an antiderivative of \(\ln(x)\). - Take the magnitude of the signed integral on the interval below the axis.

1. The graph crosses the \(x\)-axis where \(\ln(x)=0\), so split the interval at \(x=1\). 2. Use integration by parts to find an antiderivative: \(\int\ln(x)\,dx=x\ln(x)-x+C\). 3. On \(\left[\frac{1}{e},1\right]\), the graph lies below the axis. The signed integral is \([x\ln(x)-x]_{1/e}^{1}=-1+\frac{2}{e}\), so the area is \(A_1=1-\frac{2}{e}\). 4. On \([1,e]\), the graph lies above the axis, and \(A_2=[x\ln(x)-x]_{1}^{e}=1\). 5. Therefore, \(A=1-\frac{2}{e}+1=2-\frac{2}{e}\approx 1.264\) square units.

The area is \(2-\frac{2}{e}\approx 1.264\) square units.