CAL-18

Exponential integral

∫_0^x e^{kt} dt = (e^{kx} − 1)/k for k ≠ 0.

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IntegralsExponential integral

Governing equation

\int_0^x e^{kt}\,dt=\dfrac{e^{kx}-1}{k}

where

k: Rate k (—)
x: Upper limit (—)
I: Integral (—)

Lecture brief

Historical brief

Newton and Leibniz (1670s), Taylor, the fundamental theorem and the trapezoid rule are how change became a number. The lab differentiates, integrates and linearises in one variable. This sheet (CAL-18 — Exponential integral) is the form associated with Exponential integral. Working symbols: $k$ , $x$ $\rightarrow$ $I$ . The exponential is its own derivative, so the antiderivative is e^{kt}/k.

Purpose

Purpose: compute

I

from

k

x

in Calculus via

\int_0^x e^{kt}\,dt=\dfrac{e^{kx}-1}{k}

∫_0^x e^{kt} dt = (e^{kx} − 1)/k for k ≠ 0. Use it when a real calculus question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

k = 0.500\,\mathrm{—}

x = 2.000\,\mathrm{—}

, the governing relation

\int_0^x e^{kt}\,dt=\dfrac{e^{kx}-1}{k}

yields

I = 3.4366\,\mathrm{—}

. From 0 to x. k ≈ 0 falls back to x. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Rate k k0.5 —Upper limit x2 —

Outputs

Integral I3.4366 —

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Narration of this film

From 0 to x. k ≈ 0 falls back to x.

The exponential is its own derivative, so the antiderivative is e^{kt}/k.

Reading speed

Watch on YouTube