Computing integrals (1): Anti-derivative

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 So far, our discussion on the Riemann integral has been rather abstract. We know its definition and some of its properties. However, we still don't know how to calculate its value for a specific function. There are various techniques to compute integrals. We start from the most basic method based on the notion of anti-derivatives or primitive functions.



Definition (Anti-derivative, primitive function)

For a function \(f(x)\) on an open interval \(I\), a differentiable function \(F(x)\) on \(I\) is said to be an anti-derivative or primitive function of \(f(x)\) (on \(I\)) if \(F'(x) = f(x)\) holds.

Remark.   If \(F(x)\) is an anti-derivative of \(f(x)\), then for any constant \(C\), \(F(x) + C\) is also an anti-derivative of \(f(x)\). The following lemma shows these are the only anti-derivatives. □

Lemma

Let \(f(x)\) be a function on an open interval \(I\) with its anti-derivative \(F(x)\). Then any anti-derivative of \(f(x)\) is given as \(F(x) + C\) where \(C\) is a constant.

Proof. Suppose \(G(x)\) is also an anti-derivative of \(f(x)\). Then

\[(G(x) - F(x))' = f(x) - f(x) = 0\]

for all \(x\in I\). Therefore \(G(x) - F(x) = C\) (constant). ■

Remark. The collection of all anti-derivatives of \(f(x)\) is called the indefinite integral of \(f(x)\), denoted \(\int f(x)dx\). The constant term in an anti-derivative is called the constant of integration. We most often omit the constant of integration. □

Theorem

Let \(f(x)\) be a continuous function on an open interval \(I\). Then \(f(x)\) has an anti-derivative on \(I\). More precisely, with an arbitrary constant \(a\in I\), \(F(x) = \int_a^xf(t)dt\) is an anti-derivative of \(f(x)\).

Proof. Trivial from the theorem that continuous functions are integrable and the fundamental theorem of calculus. ■

Remark. In this theorem, we could choose another constant \(b\in I\) to define another anti-derivative \(G(x) = \int_b^xf(t)dt\). In fact,

\[G(x) = \int_b^xf(t)dt = \int_b^af(t)dt + \int_a^xf(t)dt = F(x) + C\]

where \(C = \int_b^af(t)dt\) is a constant. □

The following corollary shows that we can compute the integral of a function from its anti-derivative.

Corollary

Let \(f(x)\) be a continuous function on an open interval \(I\) and \(F(x)\) be an anti-derivative of \(f(x)\). For any \(a, b\in I\), we have

\[\int_a^bf(x)dx = F(b) - F(a).\]

Proof. Since we can write \(F(x) = \int_a^xf(t)dt + C\) where \(C\) is a constant,

\[F(b) - F(a) = \left(\int_a^bf(t)dt + C\right) - \left(\int_a^af(t)dt + C\right) = \int_a^bf(t)dt.\] ■

Example. Let \(F(x) = x^2\) and \(f(x) = 2x\) Then, \(F(x)\) is an anti-derivative of \(f(x)\) because \(F'(x) = (x^2)' = 2x = f(x)\). Thus, for example,

\[\int_1^22x \,dx = 2^2 - 1^2  = 3.\] □

Example. Let \(f(x) = \cos x\). Then, its anti-derivative is \(F(x) = \sin x\). Thus, for example,

\[\int_0^{\frac{\pi}{2}}\cos x\,dx = \sin\frac{\pi}{2} - \sin 0 = 1.\]

□



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