22. Riemann Sums, Integrals and the FTC

b. The Fundamental Theorem of Calculus for Integrals

2. Leibniz's Method

The \(1^\text{st}\) Fundamental Theorem of Calculus tells us how to differentiate an integral with respect to its upper limit. Leibniz's Method generalizes this using the \(2^\text{nd}\) Fundamental Theorem of Calculus. It tellls us how to differentiate an integral with respect to its lower limit or when one or both limits are functions of the variable. Leibniz's Method is actually a technique which is best learned by example.

First a simple example:

It is known that the integral \(\displaystyle \int \sin(t^2)\,dt\) cannot be expressed in terms of standard functions. However, we can still compute it's derivative. The first problem uses the \(1^\text{st}\) Fundamental Theorem of Calculus. The second problem uses the \(2^\text{nd}\) Fundamental Theorem of Calculus. The third problem combines the \(2^\text{nd}\) Fundamental Theorem of Calculus with the Chain Rule. The method in the third problem is called Leibniz's Method.

1. Compute \(\displaystyle \dfrac{d}{dx}\int_4^x \sin(t^2)\,dt\).

   By the \(1^\text{st}\) Fundamental Theorem of Calculus, the derivative of an integral with respect to its upper limit is just the integrand with the variable changed to the upper limit: \[ \dfrac{d}{dx}\int_4^x \sin(t^2)\,dt=\sin(x^2) \]

2. Compute \(\displaystyle \dfrac{d}{dx}\int_x^4 \sin(t^2)\,dt\).

   In order to use the \(2^\text{nd}\) Fundamental Theorem of Calculus, we let \(F(t)\) be any antiderivative of the integrand \(\sin(t^2)\). In other words: \[ F'(t)=\sin(t^2) \qquad \qquad (*) \] Then by the \(2^\text{nd}\) Fundamental Theorem of Calculus: \[ \int_x^4 \sin(t^2)\,dt=F(4)-F(x) \] We can now differentiate this, but first notice that \(F(4)\) is a constant. So: \[\begin{aligned} \dfrac{d}{dx}&\int_x^4 \sin(t^2)\,dt =\dfrac{d}{dx}(F(4)-F(x)) \\ &=-\,F'(x)=-\,\sin(x^2) \end{aligned}\] where in the last step we have used equation (*) with \(t\) replaced by \(x\). Thus, the derivative of an integral with respect to its lower limit is just the negative of the integrand with the variable changed to the lower limit.

   By the Second Fundamental Theorem \[ \int_b^a f(x)\,dx=F(a)-F(b) =-[F(b)-F(a)]=-\int_a^b f(x)\,dx \] Thus, in general: Reversing the limits on an integral, flips the sign of the integral. Consequently: \[\begin{aligned} \dfrac{d}{dx}&\int_x^4 \sin(t^2)\,dt =-\,\dfrac{d}{dx}&\int_4^x \sin(t^2)\,dt =-\sin(x^2) \end{aligned}\]

3. Compute \(\displaystyle \dfrac{d}{dx}\int_{3x}^{x^2} \sin(t^2)\,dt\).

   Here's where we really use Leibniz's Method: We again let \(F(t)\) be any antiderivative of the integrand \(\sin(t^2)\). So: \[ F'(t)=\sin(t^2) \qquad \qquad (*) \] Then by the \(2^\text{nd}\) Fundamental Theorem of Calculus: \[ \int_{3x}^{x^2} \sin(t^2)\,dt=F(x^2)-F(3x) \] We can now differentiate this, but we need to use the Chain Rule: \[\begin{aligned} \dfrac{d}{dx}&\int_{3x}^{x^2} \sin(t^2)\,dt =\dfrac{d}{dx}\left[F(x^2)-F(3x)\right] \\ &=F'(x^2)\dfrac{d}{dx}[x^2]-F'(3x)\dfrac{d}{dx}[3x] \\ &=2x\sin(x^4)-3\sin(9x^2) \end{aligned}\] where in the last step we have used equation (*) with \(t\) replaced first by \(x^2\) and then by \(3x\).

The Sine Integral function is defined by \[ \mathrm{Si}(x)=\int_0^x \dfrac{\sin t}{t}\,dt \] It is known that \(\mathrm{Si}(x)\) cannot be expressed in terms of standard functions. However, we can still compute its derivative.

1. Compute \(\displaystyle \dfrac{d}{dx}\mathrm{Si}(x)\).

   Use the \(1^\text{st}\) Fundamental Theorem of Calculus.

   \(\displaystyle \dfrac{d}{dx}\mathrm{Si}(x)=\dfrac{\sin x}{x}\)

   By the \(1^\text{st}\) Fundamental Theorem of Calculus, the derivative of an integral with respect to its upper limit is just the integrand with the variable changed to the upper limit: \[\begin{aligned} \dfrac{d}{dx}\mathrm{Si}(x) &=\dfrac{d}{dx}\int_0^x \dfrac{\sin t}{t}\,dt \\ &=\dfrac{\sin x}{x} \end{aligned}\]

2. Compute \(\displaystyle \dfrac{d}{dx}\int_x^\pi \dfrac{\sin t}{t}\,dt\).

   Use the \(2^\text{nd}\) Fundamental Theorem of Calculus.

   \(\displaystyle \dfrac{d}{dx}\int_x^\pi \dfrac{\sin t}{t}\,dt=-\,\dfrac{\sin x}{x}\)

   Since \(\mathrm{Si}(t)\) is an antiderivative of \(\dfrac{\sin t}{t}\), we know: \[ \mathrm{Si}'(t)=\dfrac{\sin t}{t} \qquad \qquad (*) \] Then the \(2^\text{nd}\) Fundamental Theorem of Calculus says: \[ \int_x^\pi \dfrac{\sin t}{t}\,dt=\mathrm{Si}(\pi)-\mathrm{Si}(x) \] So the derivative is: \[\begin{aligned} \dfrac{d}{dx}\int_x^\pi \dfrac{\sin t}{t}\,dt &=\dfrac{d}{dx}(\mathrm{Si}(\pi)-\mathrm{Si}(x)) \\ &=-\,\mathrm{Si}'(x)=-\,\dfrac{\sin x}{x} \end{aligned}\] where in the last step note that \(\mathrm{Si}(\pi)\) is constant and we have used equation (*) with \(t\) replaced by \(x\).

   Alternatively, we could have simply reversed the limits and multiplied by a minus.

3. Compute \(\displaystyle \dfrac{d}{dx}\int_{2\pi x}^{3\pi x} \dfrac{\sin t}{t}\,dt\).

   Combine the \(2^\text{nd}\) Fundamental Theorem of Calculus with the Chain Rule.

   \(\displaystyle \dfrac{d}{dx}\int_{2\pi x}^{3\pi x} \dfrac{\sin t}{t}\,dt =\dfrac{\sin(3\pi x)}{x}-\dfrac{\sin(2\pi x)}{x}\)

   We again observe that \(\mathrm{Si}(t)\) is an antiderivative of the integrand \(\dfrac{\sin t}{t}\): \[ \mathrm{Si}'(t)=\dfrac{\sin t}{t} \qquad \qquad (*) \] Then by the \(2^\text{nd}\) Fundamental Theorem of Calculus: \[ \int_{2\pi x}^{3\pi x} \dfrac{\sin t}{t}\,dt =\mathrm{Si}(3\pi x)-\mathrm{Si}(2\pi x) \] We can now differentiate this, using the Chain Rule: \[\begin{aligned} \dfrac{d}{dx}&\int_{2\pi x}^{3\pi x} \dfrac{\sin t}{t}\,dt =\dfrac{d}{dx}\left[\mathrm{Si}(3\pi x)-\mathrm{Si}(2\pi x)\right] \\ &=\mathrm{Si}'(3\pi x)\dfrac{d}{dx}[3\pi x]-\mathrm{Si}'(2\pi x)\dfrac{d}{dx}[2\pi x] \\ &=\dfrac{\sin(3\pi x)}{3\pi x}3\pi-\dfrac{\sin(2\pi x)}{2\pi x}2\pi =\dfrac{\sin(3\pi x)}{x}-\dfrac{\sin(2\pi x)}{x} \end{aligned}\] where in the next to last step we have used equation (*) with \(t\) replaced first by \(3\pi x\) and then by \(2\pi x\).