5.3 Fundamental Theorem of Calculus
We now show that every continuous function on \([a,b]\) is the derivative of its integral. If \(f\in R[a,b]\), then \(f\in R[a,x]\) for every \(x\in [a,b]\), so the function \[F(x) = \int ^x_af(t)\, dt,\qquad x\in [a,b],\] is well defined. It is called the indefinite integral of \(f\).
Theorem 5.3.1. Let \(f\in R[a,b]\) and \(F(x) = \int ^x_af(t)\, dt\) for \(x\in [a,b]\). Then \(F\) is continuous on \([a,b]\). If \(f\) is continuous at a point \(c\in [a,b]\), then \(F\) is differentiable at \(c\) and \(F'(c) = f(c)\).
Proof. Continuity. Let \(M\) be a bound for \(|f|\) on \([a,b]\). If \(x, x + h\in [a,b]\), then \[|F(x + h) - F(x)| = \left |\int ^{x + h}_xf(t)\, dt\right | \leq M |h| .\] So \(F\) is Lipschitz, and in particular continuous, on \([a,b]\).
Differentiability. Let \(f\) be continuous at \(c\) and \(\varepsilon > 0\). There is a \(\delta > 0\) such that \(|f(t) - f(c)| < \varepsilon \) whenever \(t\in [a,b]\) and \(|t - c| < \delta \). If \(0<|h|<\delta \) and \(c+h\in [a,b]\), then, since \(\frac {1}{h}\int _c^{c+h} f(c)\,dt = f(c)\), \[\left |\frac {F(c + h) - F(c)}{h} - f(c)\right | = \left |\frac {1}{h} \int ^{c+h}_c\bigl (f(t) - f(c)\bigr ) dt\right | \leq \frac {1}{|h|}\cdot |h|\,\varepsilon = \varepsilon .\] Hence \(F'(c) = f(c)\) (a one-sided derivative if \(c\) is an end point). □
Theorem 5.3.2 (First Fundamental Theorem of Calculus). If \(f\) is continuous on \([a,b]\) and \(F(x) = \int ^x_af(t)\, dt\), then \(F'(x) = f(x)\) for every \(x\in [a,b]\).
Proof. This follows from the previous theorem, since \(f\) is continuous at every point. Here is a second proof, using the Mean Value Theorem of Integrals. Let \(x\in [a,b]\) and \(h\neq 0\) with \(x+h\in [a,b]\). By that theorem, \begin {equation} \label {eq:ftcquot} \frac {F(x + h) - F(x)}{h} = \frac {1}{h}\int _x^{x+h}f(t)\,dt = f(C(h)) \end {equation} for some \(C(h)\) between \(x\) and \(x+h\). As \(h\longrightarrow 0\), \(C(h)\longrightarrow x\), and since \(f\) is continuous at \(x\), the right side of (5.3) tends to \(f(x)\). Hence \(F'(x) = f(x)\). □
Theorem 5.3.3 (Second Fundamental Theorem of Calculus). If \(f\) is continuous on \([a,b]\) and \(\Phi \) is a function on \([a,b]\) with \(\Phi ' (x) = f(x)\) for every \(x\in [a,b]\), then \[\int ^b_af(x)\, dx = \Phi (b) - \Phi (a) .\]
Proof. Let \(F(x) = \int ^{x}_a f(t)\, dt\). By the First Fundamental Theorem, \(F' = f = \Phi '\) on \([a,b]\), so \((F-\Phi )' = 0\), and \(F - \Phi \) is constant on \([a,b]\) (by the Mean Value Theorem). Hence \[\int _a^bf = F(b) - F(a) = \Phi (b) - \Phi (a),\] since \(F(a) = 0\). We write \(\Phi (b) - \Phi (a) = \Phi (x)\big |_a^b\). □
Example 5.3.4. Find \(\displaystyle \frac {d}{dx}\int _0^{x^2}\sin (t^2)\,dt\) and \(\displaystyle \frac {d}{dx}\int _x^{3x}\frac {dt}{t}\) for \(x>0\).
Solution. Let \(F(u) = \int _0^u\sin (t^2)\,dt\), so that \(F'(u) = \sin (u^2)\). The first function is \(F(x^2)\), and by the Chain Rule its derivative is \(F'(x^2)\cdot 2x = 2x\sin (x^4)\).
For the second, with \(G(u) = \int _1^u\frac {dt}{t}\), the function is \(G(3x) - G(x)\), with derivative \(\frac {3}{3x} - \frac {1}{x} = 0\). So \(\int _x^{3x}\frac {dt}{t}\) is constant, equal to its value \(\ln 3\) (in agreement with \(\ln 3x - \ln x = \ln 3\)).
Example 5.3.5. Evaluate \(\displaystyle \lim _{x\rightarrow 0}\frac {1}{x}\int _0^x\cos (t^2)\,dt\).
Solution. Let \(F(x) = \int _0^x\cos (t^2)\,dt\). Then \(F(0) = 0\), and by the Fundamental Theorem \(F'(0) = \cos 0 = 1\). The limit is \(\lim _{x\rightarrow 0}\frac {F(x) - F(0)}{x - 0} = F'(0) = 1\). No antiderivative of \(\cos (t^2)\) in elementary terms is needed.
Example 5.3.6. Let \(f = \operatorname {sgn}\) on \([-1,1]\) (equal to \(-1\), \(0\), \(1\) for \(x<0\), \(x=0\), \(x>0\)) and \(F(x) = \int _{-1}^xf(t)\,dt\). Find \(F\), and show that \(F\) is continuous everywhere but not differentiable at \(0\).
Solution. The function \(f\) is integrable, having one discontinuity. For \(x\leq 0\), \(F(x) = \int _{-1}^x(-1)\,dt = -(x+1)\); for \(x\geq 0\), \(F(x) = -1 + \int _0^x 1\,dt = x - 1\). So \(F(x) = |x| - 1\). This is continuous, as the first theorem of this section guarantees, and \(F'(x) = f(x)\) for \(x\neq 0\). At \(0\), where \(f\) is discontinuous, \(F\) has a corner and no derivative. So the hypothesis “\(f\) continuous at \(c\)” cannot be dropped.
Solution. The curves meet where \(x^2 = \sqrt {x}\), that is, at \(x = 0\) and \(x = 1\), and \(\sqrt {x}\geq x^2\) on \([0,1]\). By the Second Fundamental Theorem the area is \[\int _0^1\left (\sqrt {x} - x^2\right )dx = \left (\frac {2}{3}x^{3/2} - \frac {x^3}{3}\right )\Big |_0^1 = \frac {2}{3} - \frac {1}{3} = \frac {1}{3} .\]
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