Apply linearity correctly
Stacking one region on another
The figure shows two regions on the interval from 0 to 4. The teal triangle under \(f(x) = x\) has area 8, and the green rectangle under \(g(x) = 3\) has area 12. Green marks a height that is added.
Move the slider that stacks the rectangle. Each column of the rectangle moves up until its bottom sits on the line f. Its height is still 3, so its area does not change.
The stacked columns fill the region under \(f + g\), the line \(y = x + 3\), so its area is the sum of the two areas, \(8 + 12 = 20\).
Move the factor \(k\) to 2. Every column of the triangle becomes twice as tall, so its area becomes \(2\cdot 8 = 16\).
Heights add and heights scale
The curve \(f\) and the band \(g\) are now any two functions. The band is green where \(g\) is positive and red where \(g\) is negative, because a negative height takes area away.
Move the slider that stacks the band. At each \(x\) the column of \(g\) sits on the column of \(f\), so the stacked column has the height \(f(x) + g(x)\). Because the heights add at every \(x\), the area under \(f + g\) is the sum of the two areas.
Move the factor \(k\). Every column of \(f\) becomes \(k\) times as tall, so the area under \(k\,f\) becomes \(k\) times as large.
Notation
The integral of a sum is the sum of the integrals, because the heights add at every x. The bounds stay the same.
\[\begin{aligned} &\int_a^b \bigl(f(x) + g(x)\bigr)\,dx \\ &\quad = \cb{\int_a^b f(x)\,dx} + \cg{\int_a^b g(x)\,dx} \end{aligned}\]A constant factor k comes out of the integral, because every column becomes k times as tall. The faint dashed line is f before the stretch.
\[\int_a^b k\,f(x)\,dx = k\int_a^b f(x)\,dx\]The rules also work for a difference, so together they split any sum of constant multiples.
\[\begin{aligned} &\int_a^b \bigl(k f(x) - m g(x)\bigr)\,dx \\ &\quad = k\int_a^b f(x)\,dx - m\int_a^b g(x)\,dx \end{aligned}\]A constant c has the same height everywhere, so its region is a rectangle with height c and width b − a.
\[\int_a^b c\,dx = c\,(b - a)\]
When it fails
Linearity covers sums and constant multiples, but not products. Take \(f(x) = g(x) = x\) on the interval from 0 to 4, so each integral is 8.
Press the button to multiply the heights. The curve becomes \(y = x\cdot x = x^2\), above the faint dashed line \(y = x\). The region under it has area \(\frac{64}{3}\), about 21.3 as a check, but the product of the two integrals is \(8\cdot 8 = 64\). The numbers do not match, so a factor that depends on \(x\) cannot leave the integral.
Example 1: a sum with a constant
Evaluate \(\displaystyle\int_0^6 \left(5 - \dfrac{x}{2}\right)dx\), and check the result with a trapezoid.
Split the integral into the integral of 5 and the integral of \(\tfrac{x}{2}\). The red band shows \(-\tfrac{x}{2}\), below the axis, so it counts as negative.
\[\int_0^6 5\,dx - \int_0^6 \tfrac{x}{2}\,dx\]The constant 5 gives a rectangle with height 5 and width 6.
\[\int_0^6 5\,dx = 5(6) = \cb{30}\]Take the constant \(\tfrac12\) out of the second integral. The triangle under \(y = x\) from 0 to 6 has legs 6 and 6, so its area is 18, and half of it is 9. The red band cuts this 9 from the top of the rectangle.
\[\int_0^6 \tfrac{x}{2}\,dx = \tfrac12\cdot 18 = \cr{9}\]Subtract the second integral from the first. As a check, the remaining teal region is a trapezoid with parallel sides 5 and 2 and width 6, so its area is 21.
\[30 - 9 = 21 = \tfrac{5 + 2}{2}\cdot 6\]
Example 2: a constant that is not a factor
\(\displaystyle\int_1^5 f(x)\,dx = 6\). Find \(\displaystyle\int_1^5 \bigl(2f(x) + 4\bigr)\,dx\).
Split the sum, and take the constant factor 2 out of the first integral. The curve stretches to twice its height, so the first part is \(2\cdot 6 = 12\).
\[\begin{aligned} &\int_1^5 \bigl(2f(x) + 4\bigr)dx \\ &\quad = 2\int_1^5 f(x)\,dx + \int_1^5 4\,dx \\ &\quad = \cb{12} + \int_1^5 4\,dx \end{aligned}\]A tempting wrong move: The integral of 4 is 4.
\[\int_1^5 4\,dx = 4\]The constant 4 is a height, not a factor. The green band adds 4 to every column across the width of 4. A rectangle with height 4 and area 4 would be only 1 wide.
The constant 4 gives a rectangle with height 4 and width \(5 - 1 = 4\).
\[\int_1^5 4\,dx = 4(5 - 1) = \cg{16}\]Add the two parts.
\[12 + 16 = 28\]
Practice
\(\displaystyle\int_0^3 f(x)\,dx = 5\) and \(\displaystyle\int_0^3 g(x)\,dx = 2\). Find \(\displaystyle\int_0^3 \bigl(7g(x) - 2f(x) + 1\bigr)\,dx\).
Move \(k\) to 7 to stretch \(g\). Each constant factor comes out of its integral, and the constant 1 is a rectangle with width 3.
Solution
Split the integral into three integrals with the same bounds, and take the constants 7 and 2 out.
\[7\int_0^3 g\,dx - 2\int_0^3 f\,dx + \int_0^3 1\,dx\]The curve g stretches to 7 times its height, so its area becomes 7 times as large.
\[7\int_0^3 g\,dx = 7(2) = \cb{14}\]The constant 1 is a rectangle with height 1 and width 3, so its integral is 3. The red band shows \(1 - 2f\), this rectangle together with the term for \(f\), and its area \(3 - 10 = -7\) takes 7 from the region under \(7g\).
\[1(3) - 2(5) = \cr{-7}\]Add the parts.
\[\begin{aligned} 7(2) - 2(5) + 1(3) &= 14 - 10 + 3 \\ &= 7 \end{aligned}\]
Questions
The figure shows \(f(x) = 6 - x\) and \(g(x) = 2\) on the interval from 0 to 6. The region under \(f\) has area 18, and the region under \(g\) has area 12.
What is \(\displaystyle\int_0^6 \bigl(f(x) + g(x)\bigr)\,dx\)?
- 30
- 216
- 18
Answer
30
What is \(\displaystyle\int_0^6 2f(x)\,dx\)?
- 36
- 20
- 18
Answer
36
\(\displaystyle\int_1^4 f(x)\,dx = 9\). What is \(\displaystyle\int_1^4 \left(\dfrac{f(x)}{3} + 2\right)dx\)?
- 9
- 5
- 6
Answer
9
Summary
You split a sum into separate integrals, take constant factors out, and integrate a constant as a rectangle.
This works because the heights add and scale at every x, so the areas add and scale in the same way.
You use it when the integrals of f and g are known, but not for a product or a quotient of functions.
This is one of four skills in topic 6.6 of Which & Why: AP Calculus.