Integration

Answer: $4\sin(2x) + 6$

$\int \cos(2x)\,dx = \dfrac{1}{2}\sin(2x)\mathclose{,}$ so $F(x) = 4\sin(2x) + C\mathclose{.}$ At $x = 0\mathclose{,}$ $\sin 0 = 0\mathclose{:}$ $F(0) = C = 6\mathclose{.}$

Common mistakes, and the answer each one gives:

• Got the sign wrong: the antiderivative of $\cos$ is $+\sin$ → $-4\sin(2x) + 6$
• Forgot to divide by $2$ → $8\sin(2x) + 6$
• Left out the constant $C$ → $4\sin(2x)$

Answer: $\frac{18 \pi}{5}$

They meet at $x = 0$ and $x = 3\mathclose{,}$ and between them the line is on top. Each slice is a washer: $V = \pi\displaystyle\int_{0}^{3} \left(x^{2} - \left(\dfrac{x^{2}}{3}\right)^{2}\right) dx = \frac{18 \pi}{5}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Squared the difference instead of subtracting the squares → $\frac{9 \pi}{10}$
• Left out the $\pi$ → $\frac{18}{5}$

Answer: $2e^{3x} - 4$

$\int e^{3x}\,dx = \dfrac{1}{3}e^{3x}\mathclose{,}$ so $F(x) = 2e^{3x} + C\mathclose{.}$ At $x = 0\mathclose{,}$ $e^{0} = 1\mathclose{:}$ $F(0) = 2 + C = -2$ gives $C = -4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied by $3$ instead of dividing by it → $18e^{3x} - 20$
• Forgot to divide by $3$ → $6e^{3x} - 8$
• Took $C = F(0)\mathclose{,}$ forgetting that $e^{0} = 1$ → $2e^{3x} - 2$

Answer: $\dfrac{4}{3}$

They meet where $x^{2} = 6 x - 8\mathclose{,}$ that is $x^{2} - 6 x + 8 = \left(x - 2\right)\left(x - 4\right) = 0\mathclose{:}$ at $x = 2$ and $x = 4\mathclose{.}$ Between them the line is on top, so the area is $\displaystyle\int_{2}^{4} \left(- x^{2} + 6 x - 8\right) dx = \dfrac{4}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Subtracted the wrong way round, parabola minus line → $- \frac{4}{3}$
• Integrated only the parabola between the meeting points → $\frac{56}{3}$

Answer: $-\dfrac{4}{x} + 6$

$\dfrac{4}{x^{2}} = 4x^{-2}\mathclose{:}$ raise the power to $-1$ and divide by $-1\mathclose{:}$ $F(x) = -\dfrac{4}{x} + C\mathclose{.}$ $F(1) = -4 + C = 2\mathclose{,}$ so $C = 6\mathclose{.}$

Common mistakes, and the answer each one gives:

• Got the sign wrong after dividing by $-1$ → $\dfrac{4}{x} - 2$
• Lowered the power to $-3$ instead of raising it → $-\dfrac{8}{x^{3}} + 10$

Answer: $\frac{3}{2}$

Integrate term by term: $\int \left(3 x - 3\right) dx = \frac{3 x^{2}}{2} - 3 x\mathclose{.}$ Then substitute the upper limit and subtract the lower: $\left[\frac{3 x^{2}}{2} - 3 x\right]_{1}^{2} = 0 - \left(- \frac{3}{2}\right) = \frac{3}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Subtracted the wrong way round → $- \frac{3}{2}$
• Substituted the limits into the ORIGINAL function, not its integral → $3$

Answer: $\frac{81 \pi}{2}$

Each slice is a disc of radius $y\mathclose{:}$ $V = \pi\displaystyle\int_{0}^{3} y^{2} \, dx = \pi\int_{0}^{3} 9 x \, dx = \frac{81 \pi}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to square $y$ → $6 \sqrt{3} \pi$
• Left out the $\pi$ → $\frac{81}{2}$

Answer: $\frac{x^{6}}{3} - 8 x + 3$

Raise each exponent by one and divide by the new exponent: $F(x) = \dfrac{2x^{6}}{6} - 8 x + C\mathclose{.}$ Every term but $C$ is $0$ at $x = 0\mathclose{,}$ so $F(0) = C = 3\mathclose{,}$ and $F(x) = \frac{x^{6}}{3} - 8 x + 3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Differentiated instead of integrating → $10 x^{4}$
• Raised the exponent but did not divide by the new one → $2 x^{6} - 8 x + 3$
• Left out the constant $C$ → $\frac{x^{6}}{3} - 8 x$

Answer: $4\ln(x) + 2$

$\int \dfrac{1}{x}dx = \ln(x)$ for $x > 0\mathclose{,}$ so $F(x) = 4\ln(x) + C\mathclose{.}$ At $x = 1\mathclose{,}$ $\ln(1) = 0\mathclose{:}$ $F(1) = C = 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the antiderivative of $\frac{1}{x}$ to be $-\frac{1}{x^2}$ → $-\dfrac{4}{x^{2}} + 6$
• Used $\ln(1) = 1$ when finding $C$ → $4\ln(x) - 2$

Answer: $\dfrac{22}{3}$

The graph crosses the axis at $x = 1\mathclose{:}$ below it on $[0, 1]$ and above it on $[1, 3]\mathclose{.}$ An area is never negative, so integrate the two parts separately and add their sizes: $\displaystyle\int_{0}^{1} \left(1 - x^{2}\right) dx + \int_{1}^{3} \left(x^{2} - 1\right) dx = \dfrac{2}{3} + \dfrac{20}{3} = \dfrac{22}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the plain integral from $0$ to the end, so the part below the axis counted negative → $6$
• Gave only the part below the axis → $\dfrac{2}{3}$

Answer: $8 \pi$

About the $y$-axis the slices are discs of radius $x\mathclose{,}$ with $x^{2} = y\mathclose{:}$ $V = \pi\displaystyle\int_{0}^{4} x^{2} \, dy = \pi\int_{0}^{4} y \, dy = 8 \pi\mathclose{.}$

Common mistakes, and the answer each one gives:

• Rotated about the $x$-axis instead → $\frac{32 \pi}{5}$
• Left out the $\pi$ → $8$

Answer: $2x^{\frac{3}{2}} + 3$

$\sqrt{x} = x^{\frac{1}{2}}\mathclose{:}$ raise the power by one to $\frac{3}{2}$ and divide by it: $F(x) = 3 \cdot \dfrac{2}{3}x^{\frac{3}{2}} + C = 2x^{\frac{3}{2}} + C\mathclose{.}$ $F(1) = 2 + C = 5\mathclose{,}$ so $C = 3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied by $\frac{3}{2}$ instead of dividing by it → $\dfrac{9}{2}x^{\frac{3}{2}} + \dfrac{1}{2}$
• Lowered the power instead of raising it → $\dfrac{3}{2}x^{-\frac{1}{2}} + \dfrac{7}{2}$

Answer: $-2\cos(4x) - 4$

$\int \sin(4x)\,dx = -\dfrac{1}{4}\cos(4x)\mathclose{,}$ so $F(x) = -2\cos(4x) + C\mathclose{.}$ At $x = 0\mathclose{,}$ $\cos 0 = 1\mathclose{:}$ $F(0) = -2 + C = -6$ gives $C = -4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Got the sign wrong: the antiderivative of $\sin$ is $-\cos$ → $2\cos(4x) - 8$
• Forgot to divide by $4$ → $-8\cos(4x) + 2$
• Took $C = F(0)\mathclose{,}$ forgetting that $\cos 0 = 1$ → $-2\cos(4x) - 6$

Answer: $\sin(5x) + 6$

$\int \cos(5x)\,dx = \dfrac{1}{5}\sin(5x)\mathclose{,}$ so $F(x) = \sin(5x) + C\mathclose{.}$ At $x = 0\mathclose{,}$ $\sin 0 = 0\mathclose{:}$ $F(0) = C = 6\mathclose{.}$

Common mistakes, and the answer each one gives:

• Got the sign wrong: the antiderivative of $\cos$ is $+\sin$ → $-\sin(5x) + 6$
• Forgot to divide by $5$ → $5\sin(5x) + 6$
• Left out the constant $C$ → $\sin(5x)$

Answer: $- \frac{13}{4}$

Integrate term by term: $\int \left(x^{3} - 3 x^{2}\right) dx = \frac{x^{4}}{4} - x^{3}\mathclose{.}$ Then substitute the upper limit and subtract the lower: $\left[\frac{x^{4}}{4} - x^{3}\right]_{1}^{2} = -4 - \left(- \frac{3}{4}\right) = - \frac{13}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Subtracted the wrong way round → $\frac{13}{4}$
• Substituted the limits into the ORIGINAL function, not its integral → $-2$

Answer: $\dfrac{343}{6}$

They meet where $x^{2} = 12 - x\mathclose{,}$ that is $x^{2} + x - 12 = \left(x + 4\right)\left(x - 3\right) = 0\mathclose{:}$ at $x = -4$ and $x = 3\mathclose{.}$ Between them the line is on top, so the area is $\displaystyle\int_{-4}^{3} \left(- x^{2} - x + 12\right) dx = \dfrac{343}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Subtracted the wrong way round, parabola minus line → $- \frac{343}{6}$
• Integrated only the parabola between the meeting points → $\frac{91}{3}$

Answer: $\frac{64 \pi}{15}$

They meet at $x = 0$ and $x = 2\mathclose{,}$ and between them the line is on top. Each slice is a washer: $V = \pi\displaystyle\int_{0}^{2} \left((2x)^{2} - \left(x^{2}\right)^{2}\right) dx = \frac{64 \pi}{15}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Squared the difference instead of subtracting the squares → $\frac{16 \pi}{15}$
• Left out the $\pi$ → $\frac{64}{15}$

Answer: $3e^{2x} - 3$

$\int e^{2x}\,dx = \dfrac{1}{2}e^{2x}\mathclose{,}$ so $F(x) = 3e^{2x} + C\mathclose{.}$ At $x = 0\mathclose{,}$ $e^{0} = 1\mathclose{:}$ $F(0) = 3 + C = 0$ gives $C = -3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied by $2$ instead of dividing by it → $12e^{2x} - 12$
• Forgot to divide by $2$ → $6e^{2x} - 6$
• Took $C = F(0)\mathclose{,}$ forgetting that $e^{0} = 1$ → $3e^{2x}$

Answer: $-\dfrac{3}{x} - 1$

$\dfrac{3}{x^{2}} = 3x^{-2}\mathclose{:}$ raise the power to $-1$ and divide by $-1\mathclose{:}$ $F(x) = -\dfrac{3}{x} + C\mathclose{.}$ $F(1) = -3 + C = -4\mathclose{,}$ so $C = -1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Got the sign wrong after dividing by $-1$ → $\dfrac{3}{x} - 7$
• Lowered the power to $-3$ instead of raising it → $-\dfrac{6}{x^{3}} + 2$

Answer: $9 \pi$

Each slice is a disc of radius $y\mathclose{:}$ $V = \pi\displaystyle\int_{0}^{3} y^{2} \, dx = \pi\int_{0}^{3} x^{2} \, dx = 9 \pi\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to square $y$ → $\frac{9 \pi}{2}$
• Left out the $\pi$ → $9$

Answer: $\frac{x^{6}}{3} + 6 x + 2$

Raise each exponent by one and divide by the new exponent: $F(x) = \dfrac{2x^{6}}{6} + 6 x + C\mathclose{.}$ Every term but $C$ is $0$ at $x = 0\mathclose{,}$ so $F(0) = C = 2\mathclose{,}$ and $F(x) = \frac{x^{6}}{3} + 6 x + 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Differentiated instead of integrating → $10 x^{4}$
• Raised the exponent but did not divide by the new one → $2 x^{6} + 6 x + 2$
• Left out the constant $C$ → $\frac{x^{6}}{3} + 6 x$

Answer: $\dfrac{98}{3}$

The graph crosses the axis at $x = 3\mathclose{:}$ below it on $[0, 3]$ and above it on $[3, 5]\mathclose{.}$ An area is never negative, so integrate the two parts separately and add their sizes: $\displaystyle\int_{0}^{3} \left(9 - x^{2}\right) dx + \int_{3}^{5} \left(x^{2} - 9\right) dx = 18 + \dfrac{44}{3} = \dfrac{98}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the plain integral from $0$ to the end, so the part below the axis counted negative → $-\dfrac{10}{3}$
• Gave only the part below the axis → $18$

Answer: $3\ln(x) - 3$

$\int \dfrac{1}{x}dx = \ln(x)$ for $x > 0\mathclose{,}$ so $F(x) = 3\ln(x) + C\mathclose{.}$ At $x = 1\mathclose{,}$ $\ln(1) = 0\mathclose{:}$ $F(1) = C = -3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the antiderivative of $\frac{1}{x}$ to be $-\frac{1}{x^2}$ → $-\dfrac{3}{x^{2}}$
• Used $\ln(1) = 1$ when finding $C$ → $3\ln(x) - 6$

Answer: $\frac{\pi}{2}$

About the $y$-axis the slices are discs of radius $x\mathclose{,}$ with $x^{2} = y\mathclose{:}$ $V = \pi\displaystyle\int_{0}^{1} x^{2} \, dy = \pi\int_{0}^{1} y \, dy = \frac{\pi}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Rotated about the $x$-axis instead → $\frac{\pi}{5}$
• Left out the $\pi$ → $\frac{1}{2}$

Answer: $2x^{\frac{3}{2}} - 5$

$\sqrt{x} = x^{\frac{1}{2}}\mathclose{:}$ raise the power by one to $\frac{3}{2}$ and divide by it: $F(x) = 3 \cdot \dfrac{2}{3}x^{\frac{3}{2}} + C = 2x^{\frac{3}{2}} + C\mathclose{.}$ $F(1) = 2 + C = -3\mathclose{,}$ so $C = -5\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied by $\frac{3}{2}$ instead of dividing by it → $\dfrac{9}{2}x^{\frac{3}{2}} - \dfrac{15}{2}$
• Lowered the power instead of raising it → $\dfrac{3}{2}x^{-\frac{1}{2}} - \dfrac{9}{2}$

Answer: $\frac{3}{2}$

Integrate term by term: $\int \left(x^{2} - x\right) dx = \frac{x^{3}}{3} - \frac{x^{2}}{2}\mathclose{.}$ Then substitute the upper limit and subtract the lower: $\left[\frac{x^{3}}{3} - \frac{x^{2}}{2}\right]_{-1}^{2} = \frac{2}{3} - \left(- \frac{5}{6}\right) = \frac{3}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Subtracted the wrong way round → $- \frac{3}{2}$
• Substituted the limits into the ORIGINAL function, not its integral → $0$

Answer: $-4\cos(2x) + 2$

$\int \sin(2x)\,dx = -\dfrac{1}{2}\cos(2x)\mathclose{,}$ so $F(x) = -4\cos(2x) + C\mathclose{.}$ At $x = 0\mathclose{,}$ $\cos 0 = 1\mathclose{:}$ $F(0) = -4 + C = -2$ gives $C = 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Got the sign wrong: the antiderivative of $\sin$ is $-\cos$ → $4\cos(2x) - 6$
• Forgot to divide by $2$ → $-8\cos(2x) + 6$
• Took $C = F(0)\mathclose{,}$ forgetting that $\cos 0 = 1$ → $-4\cos(2x) - 2$

Answer: $\dfrac{125}{6}$

They meet where $x^{2} = 5 x\mathclose{,}$ that is $x^{2} - 5 x = x\left(x - 5\right) = 0\mathclose{:}$ at $x = 0$ and $x = 5\mathclose{.}$ Between them the line is on top, so the area is $\displaystyle\int_{0}^{5} \left(- x^{2} + 5 x\right) dx = \dfrac{125}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Subtracted the wrong way round, parabola minus line → $- \frac{125}{6}$
• Integrated only the parabola between the meeting points → $\frac{125}{3}$

Answer: $\frac{16 \pi}{15}$

They meet at $x = 0$ and $x = 2\mathclose{,}$ and between them the line is on top. Each slice is a washer: $V = \pi\displaystyle\int_{0}^{2} \left(x^{2} - \left(\dfrac{x^{2}}{2}\right)^{2}\right) dx = \frac{16 \pi}{15}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Squared the difference instead of subtracting the squares → $\frac{4 \pi}{15}$
• Left out the $\pi$ → $\frac{16}{15}$

Answer: $3\sin(2x) - 1$

$\int \cos(2x)\,dx = \dfrac{1}{2}\sin(2x)\mathclose{,}$ so $F(x) = 3\sin(2x) + C\mathclose{.}$ At $x = 0\mathclose{,}$ $\sin 0 = 0\mathclose{:}$ $F(0) = C = -1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Got the sign wrong: the antiderivative of $\cos$ is $+\sin$ → $-3\sin(2x) - 1$
• Forgot to divide by $2$ → $6\sin(2x) - 1$
• Left out the constant $C$ → $3\sin(2x)$