9709. Pure 1. Integration

Introduction to integration

Historically, integration came before differentiation and was concerned with finding the area under curves.

Isaac Newton famously linked integration and differentiation and so we now think of integration as the inverse of differentiation.  We will learn it in this form to start off with.

Polynomials: “Add one to the power and divide by the power”

Family of functions separated by a constant (why we need a constant of integration)


If \frac{dy}{dx}=x^n , then y= \frac{1}{n+1} x^{n+1} + c , n≠ -1.

If f'(x) = xn, then f(x) = \frac{1}{n+a} x^{n+1} + c , n≠ -1.


If \frac{dy}{dx} = kx^n , then y = \frac{k}{n+1} x^{n+1} + c, n ≠ -1.

Using function notation, if f'(x) = kxn, then f(x) = \frac{k}{n+1} x^{n+1} + c , n≠-1.

When integrating polynomials, apply the rule of integrations separately to each term.

Worked examples

  • Find y for:
    • \frac{dy}{dx} = x^4
    • \frac{dy}{dx} = x^-5
  • Find f(x) for:
    • f'(x) = 3x1/2
    • f'(x) = 3
  • Given that \frac{dy}{dx} = 6x + 2x^{-3} - 3x^{\frac{1}{2}}, find y.



Applied Integration

  • To find the constant of integration, c:
    • Integrate the functions;
    • Substitute the values (x,y) of a point on the curve, or the value at a given point f(x)=k, into the integrated function;
    • Solve the equation to find c.

Worked Examples

  • The curve C with equation y = f(x) passes through the point (4,5). Given that f'(x) = \frac {x^2-2}{\sqrt{x}}, find the equation of C.
  • Find the equation of the curve with the given derivative of y with respect to x that passes through the given point:
    • \frac{dy}{dx} = 3x^2 + 2x, point (2,10);
    • \frac{dy}{dx} = 4x^3 + \frac{2}{x^3} + 3, point (1,4);
    • \frac{dy}{dx} = \sqrt{x} + \frac{1}{4}x^2 , point (4,11);
    • \frac{dy}{dx} = \frac{3}{\sqrt{x}} - x , point (4,0);
  • The displacement of a particle at time t is given by the function f(t), where f(0)=0. Given that the velocity of the particle is given by f'(t) = 10 – 5t,
    • Find f(t);
    • Determine the displacement of the particle when t = 3.



Integrating (ax+b)n

We can extend the range of things we can integrate by considering “reverse differentiation” with powers of linear functions.


\int (ax+b)^n = \frac{(ax+b)^{n+1}}{a(n+1)} + c, provided n ≠ -1.

Worked Examples

  • \int 4(x-2)^3 dx
  • \int (x-2)^3 dx
  • \int 7(2x+5)^6 dx
  • \int 28(2x+5)^6 dx
  • \int 6(2x-1)^{-4} dx
  • \int \frac{1}{(2x-1)^{-4}} dx
  • \int \frac{-4}{\sqrt{1-8x}} dx
  • \int \frac{8}{\sqrt{1-8x}} dx
  • \int \frac{3}{\sqrt{5-2x}} dx



More complicated integrations (still with indefinite integrals)

In general, when integrating more difficult expressions, we need always to consider what we would need to differentiate to get our function.

(a) Differentiate (2x4-3)7

(b) If f’(x) = 64(2x4-3)6x3, then what is f(x)?

(a) Differentiate (x6+9)5/2

(b) Integrate 30x5(x+9)3/2 with respect to x



Definite Integration

An integral can be calculated between two limits. This is called as definite integral. A definite integral usually produces a value, whereas an indefinite integral always produces a function.

If f'(x) is the derivative of f(x) for all values of x in the interval [a,b] then the definite integral is defined as: \int_a^b f'(x) dx = [f(x)]_a^b = f(b) - f(a)

It is also worth noting that the integral from a to b is equal to the negative of the integral from b to a.

Also, the sum of the integral from a to b and the integral from b to c is equal to the integral from a to c.

Worked Examples

  • Evaluate:
    • \int_0^1 (x^{\frac{1}{3}} -1)^2 dx
    • \int_2^5 x^3
    • \int_1^3 x^4
    • \int_0^4 \sqrt{x}
    • \int_1^3 \frac{3}{x^2}
  • Given that P is constant and \int_1^5 (2Px+7) dx = 4P^2 , show that there are two possible values for P and find these values.



Area under curves

Historically, the primary purpose of integration was to calculate the area under curves.

The area between a positive curve, the x-axis and the lines x=a and x=b is given by: Area = \int_a^b y dx where y=f(x) is the equation of the curve.

When the area bounded by a curve and the x-axis is below the x-axis \int y dx gives a negative answer. So we need to be careful! We should draw a sketch, then calculate and sum the absolute values of the relevant areas.

We can also amend the formula to calculate the area between a curve and the y-axis.

The area between the curve x=f(y), the y-axis, and the lines y=a and y=b is given by: Area = \int_{y=a}^{y=b} x dy or \int_{y=a}^{y=b} f(y) dy

Worked Examples

  • Find the area bounded by the curve with equation y=(2-x)(3+x), the positive x-axis and the y-axis.
  • Find the area bounded by the curve with equation y=x(x+2)(x-2) and the x-axis.
  • Find the area bounded by the curve with equation x = y2 – 2y and the y-axis shown in the diagram:



Area between curves

To find the area between curves f(x) and g(x), we integrate (f(x)-g(x)) over the appropriate limits.  Note: g(x) may be a curve or a line (which is a special type of curve).

Worked Examples

The diagram below shows a sketch of the curve with equation y=x(5-x) and the line y=2x. Work out the shaded area.

Find the area of the finite region bounded by y = x2 and y = x – x2



Improper Integrals

  • There are two kinds of improper integrals we need to consider:
    • Integrals where at least one of the limits is infinite, such as \int_1^{\infty} \frac{1}{x^2 \sqrt{x}}dx or \int_{-\infty}^{-2} \frac{2}{x^5} dx ; and
    • Integrals where the function to be integrated is not defined at a point in the interval of integration. In this case, we will restrict our attention to examples where the function is not defined at one end of the interval, e.g. \int_{-2}^{0} \frac{2}{x^5} dx, where \frac{2}{x^5} is not defined at x=0.

We define the improper integral \int_a^{\infty} f(x)dx as \lim_{b\to\infty} \int_a^b f(x) dx , provided that the limit exists.

We define the improper integral \int_{- {\infty}}^b f(x)dx as \lim_{a\to - \infty} \int_a^b f(x) dx , provided that the limit exists.

When f(x) is defined for 0 < x < b, but f(x) is not defined when x=0, then the improper integral \int_0^b f(x)dx as \lim_{a\to 0^+} \int_a^b f(x) dx , provided that the limit exists.

Worked Examples

  • Evaluate \int_0^9 \frac{1}{\sqrt{x}}
  • Show that the improper integral \int_1^{\infty} \frac{1}{x^2 \sqrt{x}} dx has a value and find that value.
  • Show that only one of the following improper integrals has a finite value and find that value:
    • \int_{-2}^0 \frac{2}{x^5} dx
    • \int_{- \infty}^{-2} \frac{2}{x^5} dx



Volumes of Revolution

Rotating y = f(x) through 360 degrees around the x-axis forms a solid of revolution.

We can think of it as a stack of thin cylindrical discs, each of whose radii is a given function value.

So their total volume is the limit of all the delta’x’-width discs: V = \int_a^b \pi y^2 dx

The equivalent technique can be used to find the volume of a solid of revolution calculated around the y-axis between two y-values: V_{y-axis} = \int_p^q \pi x^2 dx

We can also calculate the volume of revolution when the area between two curves is rotated around the axis, being careful to calculate the integral of π(f2-g2) and not π(f-g)2

Worked Examples

  • The region between the curve y=x2, the x-axis and the lines x=1 and x=3 is rotated through 360º about the x-axis. Find the volume of revolution which is formed.
  • The region between the curve y=x2, the y-axis and the lines y=2 and y=5 is rotated through 360º about the y-axis. Find the volume of revolution which is formed.
  • The diagram below shows the shaded region bounded by the curve y = 16 – x2 and the line y=7. Find the volume of the solid obtained when the shaded region is rotated completely around the x-axis:

Exercise and General Exercises


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