Split and combine intervals
A tank filled in two stages
Water flows into a tank at a rate of \(2t\) liters per minute during the first 6 minutes. The graph is the line \(f(t) = 2t\). The region under it from 0 to 6 is a triangle with area 36, so 36 liters flow in.
The orange line is the split point \(c\). It cuts the region into a teal piece from 0 to \(c\) and an orange striped piece from \(c\) to 6.
At \(c = 4\), the teal piece is a triangle with area 16 and the orange piece has area 20, so the two pieces add to 36.
Move \(c\) with the slider. Each piece changes, but the sum stays 36.
A split point anywhere
The curve is now any curve above the axis, and the region runs from \(a\) to \(b\). The split point \(c\) cuts it into two pieces that share the point \(c\).
Drag c between the ends. One piece grows and the other shrinks, and their sum stays equal to the whole region.
If you know the whole region and one piece, subtract to find the other piece.
Move \(c\) to the right of \(b\). The piece from \(a\) to \(c\) now reaches past \(b\), so it covers too much. The piece from \(c\) back to \(b\) runs backwards, so it is red and removes the extra area.
Inside the ends or outside them, the two pieces add to the whole region.
Notation
The integral from \(a\) to \(b\) is the teal integral from \(a\) to \(\cy{c}\) plus the orange integral from \(\cy{c}\) to \(b\).
\[\begin{aligned} &\cb{\int_a^c f(x)\,dx} + \cy{\int_c^b f(x)\,dx} \\ &\quad = \int_a^b f(x)\,dx \end{aligned}\]The pieces share the endpoint \(\cy{c}\). They must not overlap, and they must not leave a gap.
To find a missing piece, subtract the known piece from the whole, in that order.
\[\begin{aligned} &\cy{\int_c^b f(x)\,dx} \\ &\quad = \int_a^b f(x)\,dx - \cb{\int_a^c f(x)\,dx} \end{aligned}\]The equation holds for every order of \(a\), \(b\) and \(\cy{c}\), because a piece with reversed bounds only changes its sign.
All the integrals use the same function \(f\), and \(f\) must have an integral on the whole interval.
Example 1: a missing piece
\(\displaystyle\int_1^9 f(x)\,dx = 17\) and \(\displaystyle\int_1^5 f(x)\,dx = 6\). Find \(\displaystyle\int_5^9 f(x)\,dx\).
Split the whole interval at 5. The two pieces share the point 5.
\[\int_1^9 f\,dx = \int_1^5 f\,dx + \int_5^9 f\,dx\]Substitute the known values.
\[17 = 6 + \int_5^9 f\,dx\]Subtract 6 from both sides. The orange piece is the whole minus the known piece.
\[\int_5^9 f\,dx = 17 - 6 = \cy{11}\]
Example 2: pieces that overlap
\(\displaystyle\int_2^8 f\,dx = 10\), \(\displaystyle\int_5^8 f\,dx = 4\) and \(\displaystyle\int_5^{11} f\,dx = 9\). Find \(\displaystyle\int_2^{11} f\,dx\).
The brackets above the curve show the given intervals from 2 to 8 and from 5 to 11. They overlap on the orange striped stretch from 5 to 8.
A tempting wrong move: Add the two known integrals that reach the ends.
\[\int_2^{11} f\,dx = 10 + 9 = 19\]The red bracket marks the stretch from 5 to 8, which lies under both intervals, so its integral, 4, is added twice. The whole region from 2 to 11 contains that stretch only once.
Cut the region into three pieces that meet at single points: 2 to 5, 5 to 8 and 8 to 11. The middle piece is 4, and each outer piece is a given integral minus that 4.
\[\begin{aligned} \int_2^5 f\,dx &= 10 - 4 = \cb{6} \\ \int_8^{11} f\,dx &= 9 - 4 = \cb{5} \end{aligned}\]Add the three pieces.
\[\int_2^{11} f\,dx = 6 + 4 + 5 = \cb{15}\]
Practice
\(\displaystyle\int_1^4 h(x)\,dx = 7\) and \(\displaystyle\int_1^9 h(x)\,dx = 2\). Find \(\displaystyle\int_4^9 h(x)\,dx\) and \(\displaystyle\int_9^4 h(x)\,dx\).
Drag the split point to 4. The piece you need is the whole minus the known piece.
Solution
Split the whole interval at 4. The piece from 1 to 4 is known, and the whole is 2.
Subtract the known piece from the whole. The result is negative because the red part below the axis is larger than the orange part above it.
\[\int_4^9 h\,dx = 2 - 7 = \cr{-5}\]The second integral has the bounds in reverse order, so it has the opposite sign. Swept from 9 to 4, the part below the axis counts as positive and is teal.
\[\int_9^4 h\,dx = \cb{5}\]
Questions
The figure shows the line \(f(x) = \tfrac{x}{2} + 1\) from 0 to 6. The teal bracket marks the interval from 0 to 4, with integral 8. The orange bracket marks 2 to 6, with integral 12. The integral over 0 to 6 is 15.
What is \(\displaystyle\int_4^6 f(x)\,dx\)?
- 7
- 23
- \(-7\)
Answer
7
The integrals over 0 to 4 and over 2 to 6 add to 20, but the integral over 0 to 6 is 15. Which stretch is counted twice, and what is its integral?
- The stretch from 2 to 4, with integral 5.
- The stretch from 0 to 2, with integral 3.
- The stretch from 4 to 6, with integral 7.
Answer
The stretch from 2 to 4, with integral 5.
\(\displaystyle\int_2^9 g\,dx = 12\) and \(\displaystyle\int_5^9 g\,dx = 20\). What is \(\displaystyle\int_2^5 g\,dx\)?
- \(-8\)
- 8
- 32
Answer
\(-8\)
Summary
You add the integrals of pieces that share a point, and you subtract a known piece from the whole to find a missing one.
This works because the signed area of a region is the sum of the signed areas of its pieces.
You use it when a problem gives integrals over some intervals and asks for the integral over a different one.
This is one of four skills in topic 6.6 of Which & Why: AP Calculus.