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Integration is the reverse of differentiation. If you take a function, differentiate it and then integrate the resulting expression, you will be left with the original function. Similarly, if you integrate something and then differentiate it, you will also be left the original function. While the derivative can be said to represent the gradient of the curve, the integral represents the area underneath the curve. There are two types of integral - definite integrals and indefinite integrals. A definite integral has upper and lower bounds defined. It gives an expression for the area underneath the curve between the upper and lower limits. In indefinite integral does not have its upper and lower bounds defined. Because of this, it must contain an arbitrary constant, \(C\). The diagram below demonstrates why this constant is needed. All three of the lines have the same gradient - differentiating them would give the same answer. However, there is not the same area under each of the curves( Area here referring to the area between the curve and the x axis). Clearly, the curve above the red section has a much greater area than the curve above the yellow section.

There are two parts to the fundamental theorem of calculus. The first fundamental theorem of calculus states that, if \(f\) is continuous on the closed interval \([a,b]\) and \(F\) is the indefinite integral of \(f\) on \([a,b]\), then \[\int^b_a f(x) dx =F(b)-F(a)\] The second fundamental theorem of calculus holds for \(f\), a continuous function on an open interval, \(I\) and \(a\), any point in \(I\), and states that if \(F\) is defined by \[F(x) = \int^x_af(t)dt, \] then \[ F'(x) = f(x) \] at each point in \(I\). What does this all actually mean? The first theorem says that the integral of a function between \(a\) and \(b\) is the same as the indefinite integral at a less the indefinite integral at \(b\). The second theorem basically says that the derivative of an indefinite integral of a function is the function itself. This is why integration is sometimes known as anti-differentiation.

If a function can be expressed in the form \[f(x) = x^n,\] then the indefinite integral of that function can be expressed as \[\int x^n dx = \frac{x^{n+1}}{n+1}+C.\] We can differentiate this expression to check that it is correct. \[\frac{d}{dx} (\frac{x^{n+1}}{n+1}+C) = (n+1)\frac{x^{n+1-1}}{n+1}+0=x^n \checkmark\] Notice how the constant becomes zero when it is differentiated. You will not be asked to integrate \( \it x^n\) for \( \it n=1\) at AS level.

Example 1: Find the indefinite integral of the function \[f(x) = \frac{1}{\sqrt x}.\] Answer: First, change the fraction and exponent to a surd: \[ f(x) = \frac{1}{\sqrt x} = x^{-\frac{1}{2}}\] Next, use the power rule of integration as explained on the previous page \[\int x^n dx = \frac{x^{n+1}}{n+1}+C.\] \[ \implies \int x^{-\frac{1}{2}}dx=\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C=\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C = 2 \sqrt x + C.\] Make sure not to forget the constant if the integral has no limits!

Integration is the reverse of differentiation. Therefore, since differentiation is a linear operator, so is integration. How should you use this when performing calculations? The fact that integration is a linear operator means that: \[\int f(x) +g(x) dx = \int f(x)dx + \int g(x)dx\] (And so on for as many functions as is necessary) If you calculate both of these indefinite integrals, both should have constant, as they have no limits. However, these constant are, as of yet, undefined so you can combine them into one. You can call this new constant \( C'\) (pronounced C prime) if you want, but \(C\) is usually also acceptable. This is dash has nothing to do with the convention of symbolising the derivative of \(y\) as \(y'\).

Example: Calculate the following integral: \[ \int x^2 - 4x^3 + 2 dx \] Answer: This integral requires the use of integration as a linear operator. \[\int f(x) +g(x) dx = \int f(x)dx + \int g(x)dx\] \[ \implies \int x^2 - 4x^3 + 2 dx = \int x^2 dx - \int4x^3 dx + \int 2dx\] The fact that integration is a linear operator means that we can also bring the constant outside of the integration sign. \[\implies \int x^2 dx - \int4x^3 dx + \int 2dx = \int x^2dx - 4 \int x^3dx +2\int dx \] The last integral might seem a little strange as it contains no \(x\) terms. However, it remember that \( x^0 = 1\). Now we are ready to calculate the integrand: \[\int x^2dx - 4 \int x^3dx +2\int dx \]\[=\frac{x^{2+1}}{2+1} + C' - 4 \Bigg( \frac{x^{3+1}}{3+1} + C'' \Bigg) + 2 \Bigg(\frac{x^{0+1}}{0+1}+C''' \Bigg)\] which simplifies to give: \[=\frac{x^3}{3} - \frac{4x^4}{4}+2x + C.\] Notice that all three constants have been combined into one.

If you are given the derivative of curve, you should be able to "Reverse differentiate " it in order to find what the equation that describes the curve is. However, as you have seen, this would give you a number of curves as the constant of integration would be still undefined. The constant of integration shifts the curve up or down the y-axis, depending on its sign. The diagram below shows three curves that all have the same derivative, but different constants of integration. Therefore, in order to calculate the exact function, you need the derivative and a point on the curve.

Example: A curve that passes through \((-1,1)\) has the derivative \(\frac{dy}{dx}=9x^2+6x. \) What is the equation of the curve? Answer: If we integrate the derivative, it will give the function. \[\int 9x^2+6xdx= 3x^3 +3x + C\] In order to evaluate the constant of integration, sub the given point into the equation of the curve.\[y=3x^3 +3x + C\]\[1=3(-1)^3+3(-1)+C\]\[1=-3-3+C\]\[1=C\] \[\therefore y= 3x^3 +3x + 1\]

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