Revise Integration for Cambridge 9709 (2.5)

Integration is the reverse process of differentiation. If we differentiate a function to get its gradient function, we can integrate the gradient function to get back to the original function. However, since the derivative of any constant is zero (e.g., the derivative of $x^2+5$ and $x^2-100$ are both $2x$), when we reverse the process, we lose information about the original constant term. To account for this, we add a 'constant of integration', denoted by $+c$, to every indefinite integral.

Standard Power Rule: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + c$ (for $n \neq -1$)

This rule extends to other standard functions you know how to differentiate. For functions with a linear argument, like $(ax+b)$, we apply the reverse of the chain rule: integrate the outer function as normal, then divide by the derivative of the inner function, which is simply the coefficient $a$.

Key Integrals for P2: $\int e^{ax+b} \, dx = \frac{1}{a}e^{ax+b} + c$ $\int \frac{1}{ax+b} \, dx = \frac{1}{a}\ln|ax+b| + c$ $\int \cos(ax+b) \, dx = \frac{1}{a}\sin(ax+b) + c$ $\int \sin(ax+b) \, dx = -\frac{1}{a}\cos(ax+b) + c$

A definite integral has limits of integration, $a$ (lower) and $b$ (upper). The result of a definite integral is a number that represents the signed area between the function's graph, the x-axis, and the vertical lines $x=a$ and $x=b$. To evaluate it, we first find the indefinite integral, $F(x)$, and then calculate $F(b) - F(a)$. The constant of integration $c$ is not needed as it would cancel out: $(F(b)+c) - (F(a)+c) = F(b)-F(a)$.

The Fundamental Theorem of Calculus: $\int_a^b f(x) \, dx = [F(x)]_a^b = F(b) - F(a)$

Sometimes we cannot integrate a function analytically, or we only have a set of data points from an experiment. In these cases, we can approximate the definite integral using numerical methods. The Trapezium Rule is a common method which works by dividing the area under the curve into a number of vertical strips and approximating each strip as a trapezium.

$\int_a^b y \, dx \approx \frac{h}{2}[y_0 + y_n + 2(y_1 + y_2 + ... + y_{n-1})]$, where $n$ is the number of strips and $h = \frac{b-a}{n}$ is the width of each strip.

To find the area enclosed between two curves, $y=f(x)$ and $y=g(x)$, we can think of it as the area under the upper curve minus the area under the lower curve. This simplifies to a single integral of the difference between the two functions. The first crucial step is to find the x-coordinates of the points of intersection by setting the functions equal to each other; these will become the limits of your integral.

Area $= \int_a^b (y_{upper} - y_{lower}) \, dx$