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[h]SL 5.5 Introduction to integration as anti differentiation of functions

[q] **Intro to Integration **

For the majority of operations & processes in Maths, we like to know how to do it in reverse. So here, we learn about the opposite of differentiation, known as integration/anti-differentiation.

FUNCTION → DIFFERENTIATION

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DERIVATIVE

[q]

So again, we will learn rules for integrating functions, starting with polynomials. This can be viewed as going from a function to its integral or from a derivative back to its original function.

NOTATION: The notation for the integral of \(f(x)\) uses \(\int f(x)dx\). The ‘dx’ part of it simply means with respect to \(x\).

[a]

RULE: To integrate functions in the form \(ax^n + bx^m + \dots\), we simply do the opposite of multiplying by the power, then subtracting 1 from the power, which is: adding 1 to the power, then dividing by that new power, i.e.:

\[\int ax^n dx = \frac{ax^{n+1}}{n+1}\]

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BOUNDARY CONDITION:

We will now explain the ‘+C’ at the end of the integral. If you consider that any constant disappears when differentiated, then when we are integrating, we must write ‘+C’ at the end, in case there was any constant in the original function.

[a]

FINDING \(C\):

If we want to find this mystery constant, then we must know a set of coordinates that fit the original function. We can then plug them in and solve for \(C\). We can then write out the whole function.

I will continue with the next pages shortly. Let me know if this is how you’d like it formatted!

[q]

FINDING \(C\): Here are some examples of that process:

Example 3: If \(\frac{dy}{dx} = 3x^2 + x\) and \(y = 10\) when \(x = 1\), find \(y\):

Integrate: \(y = x^3 + \frac{x^2}{2} + C\), plug in \((1, 10)\):

\[10 = (1)^3 + \frac{(1)^2}{2} + C \Rightarrow C = 8.5\]

Write the full equation: \(y = x^3 + \frac{x^2}{2} + 8.5\)

Note: They won’t tell you when you will be expected to find \(C\) as well.

[a]

DEFINITE INTEGRALS:

When we do \(\int f(x)dx\), we get a function as a result, and this is called an indefinite integral. A definite integral is, instead, in the form \(\int_a^b f(x) dx\). Here, we get a number as a result of evaluating the integral in the interval \(a \leq x \leq b\).

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This has one main use, which is that it gives us the area between \(f(x)\) and the x-axis, between \(a\) and \(b\) (see above).

The theory behind this involves splitting up this area into thin rectangles, and evaluating the limit as the width \(\to 0\). But this theory is beyond the scope of the SL course.

You will learn how to do this manually (covered in 5.11), but if it is a complicated function, in Paper 2, your GDC can find the definite integral, and therefore, the area under the curve.

[a]

In 5.11, we will also cover what happens if some of the curve is below the x-axis, and how to find the area between two curves.

Calculator Instructions:

– TI-nspire: Open CALCULATOR → [SHIFT] → Enter limits and function

– TI-84: [MATH] → [fnInt] → Enter limits and function

Example 5: Find the area enclosed by \(f(x) = e^{-\cos(2x)}\), the x-axis, y-axis, and \(x = 3\):

Need to evaluate \(\int_0^3 e^{-\cos(2x)} dx\). Use GDC, area = 3.60 units².

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