A-level Mathematics/OCR/C3/Integration

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Integration of Natural Equations[edit]

In core three you will be required to know the formulae for integrating a function involving a natural equation. The formulae for the integration of \frac{1}{x} is especially important because if you try to integrate it using the formulae that we learned in core two you will get a zero in the denominator. The formulae are:

 \int \operatorname{e}^{kx}\, dx = \frac{1}{k}\operatorname{e}^{kx} + c
 \int \frac{1}{x}\, dx = \ln \left|x\right| + c
The integral of an exponential function to any base is: \int\operatorname{a}^x\, dx = \frac{1}{\ln(\operatorname{a})}\operatorname{a}^x+C

Example[edit]

Integrate the following function 4x^3 - \frac{1}{x} + 4\operatorname{e}^{4x}.

Using the formulae we get:

\int 4x^3 - \frac{1}{x} + 4\operatorname{e}^{4x}\, dx = x^4 - \ln\left|x\right| + \operatorname{e}^{4x} + c

Integration Involving Linear Substitution[edit]

Currently we can not integrate directly a composite function, if we need to integrate a composite function we need to perform a linear substitution. To integrate the complex function f[g(x)]', we need to use the following procedure.

  1. Set u = g(x)
  2. Find the derivative of u and solve for dx
  3. In the function replace g(x) with u and dx with the result obtained in step 2.
  4. Integrate the function with respect to u.
  5. Replace u with g(x).

Example[edit]

Integrate (12x + 9)^8.

  1. u = 12x + 9
  2. \frac{du}{dx} = 12 and dx = \frac{du}{12}
  3. \int u^8 \frac{du}{12} which is equivalent to \frac{1}{12}\int u^8 du
  4. \frac{u^9}{108}
  5. \frac{(12x + 9)^9}{108} + C


Another more complex example:

Integrate  3x(6x-2)^5 .

  1. u = 6x - 2
  2. \frac{du}{dx} = 6 and dx = \frac{du}{6}
  3. \int 3xu^5 \frac{du}{6}
  4. This time, however, we've got an x term left in the equation. We can't integrate with respect to u if an x is lurking around, so we have to get rid of it.
  5. If u = 6x - 2, then x = \frac{u+2}{6}
  6. Substitute this to the integral:
  7. \int 3(\frac{u+2}{6})u^5 \frac{du}{6}
  8. Multiply out and generally clean up the integral:
  9. \int (\frac{u+2}{2})u^5 \frac{du}{6} = \frac{1}{6}\int (\frac{u^6+2u^5}{2}) du
  10. Split it into two and integrate:

\frac{1}{6}\int \frac{u^6}{2} du + \frac{1}{6} \int \frac{2u^5}{2}

= \frac{1}{6} (\frac{u^7}{14} + \frac{2u^6}{12}) + C

Volumes of Revolution[edit]

Integrals are used to find the volume of a shape that is created by rotating a line or set of lines around the x or y axis's. We can only revolve around an axis that is independent. This is also the way to prove the formulae for the volumes of such shapes as cones and spheres. The method that you will learn in this module is known as the disk method. The formulae are:

 V_x = \pi \int_{a}^{b} y^2\, dx
 V_y = \pi \int_{c}^{d} x^2\, dy

The procedure is:

  1. Square the function and integrate
  2. Input the highest value.
  3. Input the lower value.
  4. Subtract the result from the higher value from the result of the lower value. The answer has to be positive.
  5. If the curve is bounded by another curve do steps 1 to 5 and then subtract the lower curve from the higher curve. The a is the highest point at which the two curves meet and b is the lowest point at which the two curves meet.

Example One: Area between a Curve and an Axis[edit]

Find the volume of the solid obtained by rotating the line y = \frac{2}{2x + 9} around the x-axis and bounded by the line x = 6 and the y - axis.

  1. First we square the function and integrate
    1. \pi \int^6_0 \left(\frac{2}{2x + 9}\right)^2 \,dx
    2. \pi \int^6_0 \frac{4}{\left\{2x + 9\right\}^2} \,dx
    3. \pi\left [ \frac{-2}{2x + 9}\right]^6_0
  2. Then we input the highest value.
    1. \frac{-2}{2\times6 + 9} = \frac{-2}{21}
  3. Then we input the lowest value.
    1. \frac{-2}{2\times0 + 9} = \frac{-2}{9}
  4. Finally we subtract the result from the higher value from the result of the lower value. The answer has to be positive.
    1. \pi \left(\frac{-2}{21}- \frac{-2}{9}\right) = \frac{8}{63}\pi
  5. The area of the solid is \frac{8}{63}\pi.

Example Two: Area between Two Curves[edit]

Find the volume of the object bounded by the curves y = x^2\, and y = \sqrt{x} revolved around the x-axis.

  1. First we find where they are equal.
    1. x^2 = \sqrt{x} x = 1 or 0
  2. Then we integrate the first function.
    1. \pi \int^1_0 x^4 \, dx
    2. \pi\left[\frac{x^5}{5}\right]^1_0
  3. Now we input the higher number.
    1. \frac{1^5}{5} = \frac{1}{5}
  4. Then we input the lower number.
    1. \frac{0^5}{5} = \frac{0}{5}
  5. We subtract and we get \frac{1\pi}{5}
  6. Now we need to do the same for the second curve.
  7. We integrate the second function.
    1. \pi\int^1_0 x \, dx
    2. \pi\left[\frac{x^2}{2}\right]^1_0
  8. Now we input the higher number.
    1. \frac{1^2}{2} = \frac{1}{2}
  9. Then we input the lower number.
    1. \frac{0^2}{2} = \frac{0}{2}
  10. We subtract and we get \frac{1\pi}{2}
  11. Finally we subtract the area the upper curve from the area of the lower curve
    1. \frac{1\pi}{2} - \frac{1\pi}{5} = \frac{3\pi}{10}
  12. The area of the solid bounded by the curves y = x^2\, and y = \sqrt{x} and rotated around the x-axis is \frac{3\pi}{10}


This is part of the C3 (Core Mathematics 3) module of the A-level Mathematics text.