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Digital Lessons
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Definite Integrals

We have shown that, for example, the area under the curve \(x^2\) is given by

\[\int x^2\, dx = \frac{x^3}{3}+C\]

As with derivatives, we see that such indefinite integrals can depend on \(x\) (as well as a constant). We explained why this was the case with derivatives – the slope of a tangent to a curve is different depending on where that tangent is located on the curve.

But why do integrals depends on \(x\)? Integrals just refer to areas after all. Shouldn’t an area just be a number? Again, you’d be right!

However, notice that at no point did we mention the end points of our area. We have not defined over which \(x\) values we wish to calculate this area which explains why the area does not currently have a fixed value!

Figure 1

The size of the area depends on the values of these endpoints, i.e. on the value of \(x\), which is why it still appears.

However, we can of course specify the portion of the curve over which we want to find the area.

If we wish to calculate the area under the curve from \(x=a\) to \(x=b\)

Figure 2

we can instead calculate the definite integral given by:

\[\int_a^b y(x)\,dx\]

This is calculated as follows.

Key Point

Let \(I(x)\) be the indefinite integral of the function \(y(x)\), i.e.

\[I(x) = \int y(x)\,dx\]

The definite integral of this function between \(x=a\) and \(x=b\) is then given by

\[\int_a^b y(x)\,dx=I(b)-I(a)\]

For example, let us calculate:

\[\int_4^7 x^2\,dx\]

Figure 3

To do so, we first calculate the indefinite integral

\begin{align} I(x) &=\int x^2\,dx \\&= \frac{x^3}{3} +C \end{align}

before then calculating the definite integral

\begin{align}\int_4^7 x^2\,dx&=I(7)-I(4)\\&= \left(\frac{7^3}{3} +C\right)  – \left( \frac{4^3}{3} +C \right) \\&= \frac{7^3}{3}-\frac{4^3}{3}\\&=93\end{align}

Therefore, the area under the curve \(y(x)=x^2\) between \(x=3\) and \(x=7\) is \(93\).

Note also that the constant of integration \(C\) has disappeared. In fact, it will always disappear when dealing with definite integrals and can simply be ignored. (However, it must still be included when instead dealing with indefinite integrals!)

We can also perform a definite integral in one step by using vertical bars to represent our endpoints, as is done in the example below.

Guided Example 1

Calculate the following definite integrals

(a) \(\displaystyle\int_2^5 x^3 \, dx\)

(b) \(\displaystyle\int_{-2}^3 x^5 \, dx\)

(c) \(\displaystyle\int_0^{1} e^{2x} \, dx\)

(d) \(\displaystyle\int_\pi^{2\pi} dx\)


\begin{align}\int_2^5 x^3 \,dx & = \left.\frac{x^4}{4}\right|_2^5\\&=\frac{5^4}{4} – \frac{2^4}{4}\\&=152.25\end{align}


\begin{align}\int_{-2}^3 x^5 \, dx & = \left.\frac{x^6}{6}\right|_{-2}^3\\&=\frac{3^6}{6} – \frac{(-2)^6}{6}\\&=\frac{665}{6}\end{align}


\begin{align}\int_0^{1} e^{2x} \, dx & = \left.\frac{1}{2}e^{2x}\right|_0^{1}\\&=\frac{1}{2}e^{2(1)}-\frac{1}{2}e^{2(0)}\\&=\frac{e^2-1}{2}\end{align}


\begin{align}\int_\pi^{2\pi} dx & = \left.x\right|_\pi^{2\pi}\\&=2 \pi – \pi\\&=\pi\end{align}