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Revision:FSMQ Integration

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > FSMQ Integration

Say we are told $\frac{dy}{dx} = 3x^2$ how would we find what equals? Integration provides us with a way to move from differential equations back to original equations, as well as a way for calculating areas on graphs.

1 The Rule of Integration
2 The Constant of Integration
3 Specified Cases in Integration
4 Definite Integrals

The Rule of Integration

The general formula for integration is:

if $\frac{dy}{dx} = kx^n$ then $y = \frac{kx^{n+1}}{n+1} + c$

is known as the constant of integration. This shows that the equation we have formed is a general equation, or family of curves, as with insufficient information, we cannot know the exact equation of the curve.

For example, if $\frac{dy}{dx} = 3x^2$

the way we integrate is:

$\int 3x^2 \, dx = \frac{3x^{2+1}}{2+1} + c$

The sign for integration is shaped similar to an 'S,' this is because integration was originally used for summations. The means with respect to , so that we know what we are integrating.

Continuing the integration, we find that $\int 3x^2 \, dx = x^3 + c$

Another example: $\frac{dy}{dx} = 4x^3 + 4x^2 + 1$

$\int (4x^3 + 4x^2 + 1) \, dx = x^4 + 2x^2 + x + c$

If we are integrating more than one term, we should put a bracket around the terms being integrated, so we know that applies to all of them.

The Constant of Integration

The constant of integration, labelled , is necessary to show that we have formed a general equation, which is true for a family of curves. Here is why:

Differentiating these three functions:

y = x^3 + 7 , y = x^3 and y = x^3 - 6

We find that they all have the differential equation:

$\frac{dy}{dx} = 3x^2$

Now say the question is to integrate $\frac{dy}{dx} = 3x^2$ and you were given no other information:

We could only show that $\int 3x^2 \, dx = x^3 + c$ and we can solve it no further. So shows that this equation is general and represents a whole family of curves, which have the differential equation of 3x^2

Specified Cases in Integration

Say we are told that $\frac{dy}{dx} = 2x$

By integration we find that:

y = x^2 + c

and then we are told the graph passes through the point (2,6)

We can then find the fully equation of the graph, by substituting in x = 2 and y = 6 to find .

So in this situation:

6 = 4 + c

so c = 2

and y = x^2 + 2

Another example:

$\frac{dy}{dx} = 4x + 1$

integrating, we get, y = 2x^2 + x + c

then we are told that the graph passes through (2,13)

we find that c = 3

and that the equation is y = 2x^2 + x + 3

Definite Integrals

Definite Integrals take the form:

$\int_a^b f'(x)$

and are defined as:

$\int_a^b f'(x) \, dx = f(b) - f(a)$

We call and the limits of the integral, with as the upper limit and as the lower limit.

We can evaluate definite integrals, because when we subtract the two functions the constant of integration cancels out, and so there is no need for us to find/calculate it.

For example:

Evaluate $\int_1^4 (2x) \, dx$

we can find that f(x) = x^2 + c

so evaluating: f(4) - f(1) = (4^2 + c) - (1^2 + c) = 16 - 1 = 15

Area Under a Curve

Using limits and equations of curves, we can find the area bordered by the curve, the x-axis and the limits. Like in the above example, the area between the curve y = 2x the lines x = 4, 1 and the x-axis, is 15 units squared.

For example, if we wanted to find the area of then curve y = 3x^2 + 1 between 2 and 1:

$\int_1^2 (3x^2 + 1) \, dx$ so f(x) = x^3 + x + c

f(b) - f(a) = f(2) - f(1) = (2^2 + 2) - (1^3 + 1) = (4 + 2) - (1 + 1) = 6 - 2 = 4 units squared.

Note how we have omitted as it simply cancels out.

Area Between a Two Curves

If we wanted to find the area between y = 4x^3 + 1 and y = 2x between the x-coordinates 3 and 1:

All we need to do is integrate (f_1(x) - f_2(x))

So we get:

$\int_1^3 ((4x^3 + 1) - (2x)) \, dx$

we get then have g(3) - g(1) where g(x) = x^4 - x^2 + x + c

which evaluates as: (81 - 9 + 3) - (1-1+1) = 75 - 1 = 74 units squared.

Negative Definite Integral

If we find ourselves in a situation where the integral has been evaluated as negative, when not dealing with area, we simply leave it as negative, and not worry about it. However, when we are dealing with area, and it is negative, we cannot give an answer stating '-n units squared' and we must make it positive. If it is negative, all it shows is that the area is above the curve which we made the subject; when dealing with the area between the curve and the axis, this simply means that the area is above the curve, but when dealing with two curves, it means that the area is above the f_1(x) curve where we integrate (f_1(x) - f_2(x)) and below the f_2(x) curve.