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The Definite Integral

Expert reviewed • 21 July 2024 • 8 minute read

HSC Maths Advanced Syllabus

  • recognise and use linearity of anti-differentiation
    • examine families of anti-derivatives of a given function graphically
  • understand the relationship of position to signed areas, namely that the signed area above the horizontal axis is positive and the signed area below the horizontal axis is negative
  • using technology or otherwise, investigate the link between the anti-derivative and the area under a curve
    • interpret ∫abf(x)dx\int_a^bf(x)dx as a sum of signed areas
    • understand the concept of the signed area function F(x)=∫axf(t)dtF(x)=\int_a^x f(t)dt

Note:

Video coming soon!

What is the Definite Integral?

As we know the definite integral is an integral which calculates the accumulation of quantities, such as areas under curves, over a specific interval. For example, a definite integral can be governed by the interval [a,b][a,b]. In previous chapters, we have explored expressions using simple definite integrals, usually located in the positive y-axis. But what happens if we are to integrate with negative values?

The Signed Area Function

Discussed in the last topic, the signed area function which is the sum of signed area’s underneath a graphs curve, is part of the fundamental formula of calculus, to which its formula is:

F(x)=∫axf(t)dtF(x)=\int_a^xf(t)dt

As such, when we integrate, a negative answer represents the area between a curve and the axis, located in the negative axis, while a positive value has the opposite effect.

For example, the graph below, when integrated results in a value of −4-4. This means that the area between the graph and the x-axis, is equal to 4, and it is located below the x-axis.

placeholder

Types of Definite Integrals

Like any type of function/formula there are multiple variations which can be considered and applied depending on circumstances. For definite integrals, the following formulas/rules can be examined and used to easily integrate given functions:

  • For intervals with zero width the integral will equal to zero
∫aaf(x)dx=0\int_a^af(x)dx=0 placeholder
  • When integrating odd functions, the following applies as the function is not symmetrical
∫−aaf(x)dx=0\int_{-a}^af(x)dx=0
  • When integrating even functions, we can manipulate the following expression, due to its symmetrical nature
∫−aaf(x)dx=2∫0af(x)dx\int_{-a}^af(x)dx=2\int_0^af(x)dx
  • When a region under the graph has been dissected by another point, the following expression can be used
∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx placeholder
  • In order to reverse the interval, the integral expression must be manipulated as follows:
∫abf(x)dx=−∫baf(x)dx\int_a^bf(x)dx=-\int_b^af(x)dx
  • When an integral contains a sum of two different functions, they can be split into two different expressions:
∫ab(f(x)+g(x))dx=∫abf(x)dx+∫abg(x)dx\int_a^b(f(x)+g(x))dx=\int_a^bf(x)dx+\int_a^bg(x)dx
  • When an integral contains a multiple of a number we can simplify the expression to make calculations simpler to conduct
∫abkf(x)dx=k∫abf(x)dx\int_a^bkf(x)dx=k\int_a^bf(x)dx

Practice Question 1

Using rules from above, evaluate the following: ∫−22−x2−4.dx\int_{-2}^2-x^2-4.dx

To evaluate this integral, we can first simplify it by applying given rules above, to ensure simple calculations.

∫−22−x2−4.dx=∫−22−(x2+4)dx=−2∫02(x2+4)dx\int_{-2}^2-x^2-4.dx=\int_{-2}^2-(x^2+4)dx\\=-2\int_0^2(x^2+4)dx

Now we have simplified the expression, we can integrate and evaluate.

−2∫02(x2+4)dx=−2[12+1x2+1+40+1x0+1]02=−2[x33+4x]02=−2[(233+4(2))−(033+4(0))]=−2(83+8−0)=−643-2\int_0^2(x^2+4)dx=-2[\frac{1}{2+1}x^{2+1}+\frac{4}{0+1}x^{0+1}]_0^2\\=-2[\frac{x^3}{3}+4x]_0^2\\=-2[(\frac{2^3}{3}+4(2))-(\frac{0^3}{3}+4(0))]\\=-2(\frac{8}{3}+8-0)\\=-\frac{64}{3}

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The Fundamental Theoerem of Calculus

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The Indefinite Integral

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