The definite integral and the area to the x-axis

The definite integral of $\displaystyle f$ from $\displaystyle a$ to $\displaystyle b$ is a number: the difference between the values of an antiderivative $\displaystyle F$ at the two ends.

• $\displaystyle a$ – the lower limit, $\displaystyle b$ – the upper limit;
• the step in the middle is written $\displaystyle {\left[F(x)\right]_{a}^{b}}={F(b)-F(a)}$;
• the constant $\displaystyle C$ does not change the result: it is added to $\displaystyle F(b)$ and to $\displaystyle F(a)$, and the subtraction removes it. So $\displaystyle {C}={0}$ is taken.

In the example below we work out $\displaystyle \int_0^3(x^2+1)\,dx$:

When $\displaystyle f(x)\geq0$ between $\displaystyle a$ and $\displaystyle b$, the definite integral equals the area between the graph of $\displaystyle f$, the $\displaystyle x$-axis and the lines $\displaystyle {x}={a}$ and $\displaystyle {x}={b}$.
On the board – the area between the graph of $\displaystyle {f(x)}={x^2+1}$, the $\displaystyle x$-axis, $\displaystyle {x}={0}$ and $\displaystyle {x}={3}$. By the example, the area is 12.

When $\displaystyle f(x)\leq0$ between $\displaystyle a$ and $\displaystyle b$, the integral is negative: every value of $\displaystyle f$ is negative. The area is minus the integral.
On the board – the area between the graph of $\displaystyle {f(x)}={x^2-4}$ and the $\displaystyle x$-axis, from $\displaystyle {x}={0}$ to $\displaystyle {x}={2}$:

The area: $\displaystyle \frac{16}{3}$.

When $\displaystyle f$ changes sign between $\displaystyle a$ and $\displaystyle b$, the interval is split at the zeros of $\displaystyle f$, and the area of each part is worked out on its own.
On the board – $\displaystyle {f(x)}={x^2-2x}$ from $\displaystyle {x}={0}$ to $\displaystyle {x}={3}$. The zeros: $\displaystyle {x}={0}$, $\displaystyle {x}={2}$.

• From 0 to 2 the graph is below the axis: $\displaystyle {\int_0^2(x^2-2x)\,dx}={\left[\frac{x^3}{3}-x^2\right]_0^2}={\frac{8}{3}-4}={-\frac{4}{3}}$, and the area is $\displaystyle \frac{4}{3}$;
• from 2 to 3 the graph is above the axis: $\displaystyle {\int_2^3(x^2-2x)\,dx}={0-\left(-\frac{4}{3}\right)}={\frac{4}{3}}$;
• the whole area: $\displaystyle {\frac{4}{3}+\frac{4}{3}}={\frac{8}{3}}$.

The integral from 0 to 3 is $\displaystyle {-\frac{4}{3}+\frac{4}{3}}={0}$: the part below the axis cancels the part above it. So one integral over the whole interval does not give the area.

Work out the integrals.

On the board – the graph of $\displaystyle {f(x)}={3+2x-x^2}$, and the area it encloses with the $\displaystyle x$-axis.

Find where the graph meets the $\displaystyle x$-axis, and mark the points on the board.

Work out the area.

On the board – the graph of $\displaystyle {f(x)}={x^3-x^2-2x}$, and the two areas it encloses with the $\displaystyle x$-axis.

Find where the graph meets the $\displaystyle x$-axis.

Work out the integral over the left part.

Work out the integral over the right part.

Work out the whole area.

Work out the area bounded by the graph of $\displaystyle {f(x)}={\frac{1}{x^2}}$, the $\displaystyle x$-axis and the lines $\displaystyle {x}={1}$ and $\displaystyle {x}={2}$.