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AI HL · AHL 3

Further Integration

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AHL 3.1

Integrating powers and standard functions

AHL 3.1

Integrating powers and standard functions

Target: I can integrate powers of \(x\), including fractional and negative powers, after rewriting roots and fractions as powers. (+3 more)

By the end of this section

Key points

Powers of x

  • \(\int x^n\,dx=\frac{x^{n+1}}{n+1}+c\) for any n except \(-1\). It works for fractions and negatives too.
  • Rewrite first: \(\sqrt{x}=x^{\frac12}\), \(\sqrt[3]{x}=x^{\frac13}\), \(\frac{1}{\sqrt{x}}=x^{-\frac12}\), \(\frac{1}{x^2}=x^{-2}\).
  • Add 1 to the power, then divide by the new power. Dividing by a fraction means multiplying by its flip.
  • Do not subtract 1. That is differentiating the wrong way round.

The standard integrals

  • \(\int \cos x\,dx=\sin x+c\) and \(\int \sin x\,dx=-\cos x+c\). Angles are in radians.
  • \(\int e^x\,dx=e^x+c\) and \(\int \frac1x\,dx=\ln|x|+c\).
  • The cycle \(\sin x,\ \cos x,\ -\sin x,\ -\cos x\) and back to \(\sin x\): moving forward is differentiating, moving back is integrating.
  • The power rule never gives \(\frac1x\): that is the case \(n=-1\), and the answer is a logarithm.

Constants, sums and finding y

  • A constant multiplier stays out front: \(\int 3e^x\,dx=3e^x+c\).
  • Split a sum into separate integrals, integrate each, and write one \(+c\) at the end.
  • To find \(y\) from \(\frac{dy}{dx}\), integrate, then put in a point on the curve to find \(c\).
  • Check by differentiating your answer. You should get back the original function.

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
A fractional power

Find \(\int 6x^{\frac{2}{3}}\,dx\). Give the coefficient of \(x^{5/3}\) as a decimal.

Working
  1. Add 1 to the power

    \(\frac23+1=\frac53\)

    The power goes up by one.

  2. Divide by the new power

    \(6\div\frac53=\frac{18}{5}\)

    Dividing by a fraction: multiply by its flip, \(6\times\frac35\).

  3. Write the answer

    \(\int 6x^{\frac23}\,dx=\frac{18}{5}x^{\frac{5}{3}} + c\)

    Add c. There is no limit, so c stays.

Video walkthrough coming
A root with a fraction in front

Find \(\int \frac45\sqrt[4]{x}\,dx\).

Working
  1. Root as a power

    \(\sqrt[4]{x}=x^{\frac14}\)

    Rewrite before you integrate. The fraction 4/5 is a constant and stays out front.

  2. Add 1 and divide

    \(\frac14+1=\frac54,\quad \frac45\div\frac54=\frac45\times\frac45=\frac{16}{25}\)

    Flip 5/4 to multiply.

  3. Answer

    \(\int\frac45\sqrt[4]{x}\,dx=\frac{16}{25}x^{\frac54}+c\)

    Keep the power as a fraction or write it with a root.

Video walkthrough coming
A negative fractional power

Find \(\int 5x^{-\frac{3}{2}}\,dx\).

Working
  1. Add 1

    \(-\frac32+1=-\frac12\)

    Adding 1 to a negative power moves it towards 0.

  2. Divide

    \(5\div\left(-\frac12\right)=-10\)

    Keep the negative sign.

  3. Answer

    \(\int 5x^{-\frac32}\,dx=-10x^{-\frac12}+c\)

    You can also write \(-\frac{10}{\sqrt{x}}\).

Video walkthrough coming
A constant over a root

Find \(\int \frac{8}{\sqrt[3]{x}}\,dx\).

Working
  1. Negative power

    \(\frac{8}{\sqrt[3]{x}}=8x^{-\frac13}\)

    A root underneath is a negative power. Write it before you integrate.

  2. Add 1 and divide

    \(-\frac13+1=\frac23,\quad 8\div\frac23=12\)

    8 times 3/2.

  3. Answer

    \(\int\frac{8}{\sqrt[3]{x}}\,dx=12x^{\frac23}+c\)

    Positive power again.

Video walkthrough coming
A sum, then the constant

A curve has \(\frac{dy}{dx}=4e^x-2\sin x+3\) and passes through the point \((0,\,5)\). Find the constant c in \(y=\dots+c\).

Working
  1. Integrate each term

    \(y=4e^x+2\cos x+3x+c\)

    The integral of \(\sin x\) is \(-\cos x\), so \(-2\sin x\) gives \(+2\cos x\). One c for the whole sum.

  2. Put in the point

    \(5=4e^0+2\cos 0+3(0)+c=4+2+c\)

    \(e^0=1\) and \(\cos 0=1\).

  3. Solve for c

    \(c=-1\)

    So \(y=4e^x+2\cos x+3x-1\).

Video walkthrough coming
1/x gives a logarithm

\(\frac{dy}{dx}=\frac6x\) for \(x>0\), and \(y=10\) when \(x=e^2\). Find the constant c.

Working
  1. Integrate

    \(y=6\ln|x|+c\)

    The integral of 1/x is \(\ln|x|\). The 6 comes out front.

  2. Put in the point

    \(10=6\ln e^2+c=12+c\)

    \(\ln e^2=2\), since ln undoes e to the power.

  3. Solve

    \(c=-2\)

    So \(y=6\ln x-2\) for \(x>0\).

Video walkthrough coming