Integration

Integration is differentiation run backwards

Differentiation takes a function and produces its gradient function. Integration takes a gradient function and gives back the original — which is why it is also called antidifferentiation.

If $\dfrac{dy}{dx} = 2x$, then $y = x^2$ works. So does $y = x^2 + 7$, and $y = x^2 - 41$. Every one of them has gradient $2x$, because differentiating a constant gives zero and destroys the information. Integration cannot recover what was destroyed, so it admits as much:

$$\int 2x \, dx = x^2 + c$$

That $+c$ is the constant of integration. It is not decoration and it is not optional.

The power rule for integration

Add one to the index, then divide by the new index. That is the exact reverse of the differentiation rule, where you multiplied by the index and then subtracted one.

As with differentiation, rewrite before you integrate: $\dfrac{4}{x^3}$ becomes $4x^{-3}$, and $\sqrt{x}$ becomes $x^{1/2}$.

The condition $n \neq -1$ is there because $n + 1$ would be zero and you cannot divide by zero. Integrating $x^{-1}$ needs a logarithm, which is beyond this syllabus.

Definite integrals

A definite integral has limits, and it evaluates to a number rather than a function:

$$\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a)$$

Integrate as usual, put the result in square brackets with the limits, then substitute the top limit and subtract the value at the bottom limit.

Finding a curve from its gradient

If you are told $\dfrac{dy}{dx}$ and one point the curve passes through, you can pin down $c$ and recover the curve exactly.

1. Integrate to get $y$ in terms of $x$, with $+c$.
2. Substitute the coordinates of the given point.
3. Solve for $c$ and write out the full equation.

Area under a curve

The definite integral $\displaystyle\int_a^b y \, dx$ gives the signed area between the curve and the $x$-axis, from $x = a$ to $x = b$.

Signed means: area above the axis counts as positive, area below counts as negative.

Area between two graphs

For the region enclosed between two graphs between their points of intersection:

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

where $a$ and $b$ come from solving the two equations simultaneously.

The subtraction is what makes this work — it removes the strip beneath the lower graph, and it also means you do not have to worry about whether the region sits above or below the $x$-axis.

Kinematics

Integration reverses the chain from displacement to velocity to acceleration:

$$a \;\xrightarrow{\;\int\;}\; v \;\xrightarrow{\;\int\;}\; s$$

so $v = \displaystyle\int a\,dt$ and $s = \displaystyle\int v\,dt$.

Each integration brings a constant, and each constant is found from a condition given in the question. The two that appear most often are "starts from rest", meaning $v = 0$ when $t = 0$, and "starts at the origin", meaning $s = 0$ when $t = 0$.