Integrating exponential and logarithmic functions · Grade 12 · 4 units
In Integrating exponential and logarithmic functions (Grade 12 · 4 units) you practise 6 skills: Antiderivatives · An antiderivative through a point · The definite integral · The area between a graph and the axis · The area between two graphs · Finding a parameter. Four new integrals join the list here. The integral of e to the power of x is e to the power of x, plus a constant. The integral of 1 over x is ln x, plus a constant, when x is positive. When the power or the denominator is a linear expression, such as 2x minus 1, divide by the coefficient of x as well: the integral of e to the power of 2x is half of e to the power of 2x, and the integral of 1 over the expression 2x minus 1 is half the ln of that expression.
In a definite integral, put the bounds into the antiderivative. For example, take the integral of e to the power of x from 0 to ln 3. The antiderivative is e to the power of x, and at the upper bound it equals 3, because e to the power of ln 3 is 3. At the lower bound it equals 1. So the integral is 3 minus 1, which is 2. When the graph is below the x axis, the integral comes out negative. An area is always positive, so take the negative of the integral. When the graph crosses the axis, split the interval where it crosses and work out each part on its own. The area between two graphs is the integral of the upper graph minus the lower graph, between the points where they meet or between the given lines. If the graphs swap places along the way, split the interval and add the areas.
Tip: A power of e whose index is the ln of a number equals that number, so bounds with ln give simple numbers.
What you learn
Antiderivatives
For example: Here is the integral ∫e^(−4x) dx . Which expression is equal to it?
- (1 ÷ 4) × e^(−4x) + C
- −4 × e^(−4x) + C
- e^(−4x) + C
- −(1 ÷ 4) × e^(−4x) + C
Answer and solution
The answer: −(1 ÷ 4) × e^(−4x) + C
Divide by the coefficient of x in the power, which is −4 : −(1 ÷ 4) × e^(−4x) + C .
An antiderivative through a point
For example: The derivative of the function F is F′(x) = eˣ , and F(0) = 9 . What is the constant C in the antiderivative F(x) = eˣ + C ?
Answer and solution
The answer: 8
The integral is F(x) = eˣ + C . Put in x = 0 : e⁰ + C = 9 , that is 1 + C = 9 , so C = 8 .
The definite integral
For example: What is the value of the integral ∫₀¹ eˣ dx ?
- e
- e + 1
- 1 − e
- e − 1
Answer and solution
The answer: e − 1
The antiderivative is eˣ , so the value is e − e⁰ = e − 1 .
The area between a graph and the axis
For example: What is the area between the graph of y = 4 ÷ x , the horizontal axis and the lines x = 1 and x = e ?
Answer and solution
The answer: 4
The antiderivative is 4 × ln x , so the integral is ∫₁ᵉ (4 ÷ x) dx = 4 . That is the area, because the graph is above the axis.
The area between two graphs
For example: What is the area enclosed by the graphs y = e^(2x) , y = eˣ and the line x = ln 2 ?
Answer and solution
The answer: 1/2
The graphs meet at x = 0 , and from there the graph of e^(2x) is the upper one. The area is ∫₀^(ln 2) (e^(2x) − eˣ) dx , that is (4 ÷ 2 − 2) − (1 ÷ 2 − 1) , which is 1/2 .
Finding a parameter
For example: It is known that ∫₀^(ln k) eˣ dx = 3 , where k > 0 . What is the value of k ?
Answer and solution
The answer: 4
The integral is e^(ln k) − e⁰ = k − 1 , so k − 1 = 3 , that is k = 4 .
Not on the site yet
Parts of this topic that are not built yet: The integral of ln x and of products with e.
The exercises are made by rules and the answers are worked out in code, so there are new exercises every time. The topic follows the Israeli Ministry of Education’s high school maths curriculum at 4 units (grades 10 to 12).