Use endpoint and integral data

Only some values are known

A test question gives you no formula for a function \(f\). It gives three facts: \(f(0) = 1\), \(f(4) = 6\), and the area under the graph of \(f\) from 0 to 4 is 15. You must find \(\int_0^4 x f'(x)\,dx\).

This is \(\int u\,dv\) with \(u = x\) and \(v = f(x)\). The curve is one graph that fits the facts, drawn up to the point you can drag. The teal region under the curve and the green region beside it fill the orange rectangle, which reaches from the origin to that point.

Drag the end of the curve to \(x = 4\). The rectangle is now 4 wide and \(f(4) = 6\) tall, so its area is 24. The teal region is the given area, 15. The green region is \(\int u\,dv\), the integral you want, so its area is \(24 - 15 = 9\).

You needed no formula for f, only the values at the limits and the given area.

Values at the limits, and one given integral

The parts formula uses the values of \(u\) and \(v\) at the two limits, and one integral, \(\int v\,du\). A question can give all of these numbers instead of a formula.

The large orange rectangle is the product \(uv\) at the upper limit, and the small dashed rectangle is the product at the lower limit. The product term \(\big[uv\big]_a^b\) is the large rectangle minus the small one.

The teal region is the integral that the question gives, so the green region is the product term minus the teal region.

Press Bend the curve. The curve changes shape between its ends, but the two rectangles and the teal area stay the same, so the green area stays the same.

Notation

1. Write the integral that you want as \(\int_a^b u\,dv\). If a derivative \(f'(x)\) appears, take \(dv = f'(x)\,dx\), so \(v = f(x)\).

   \[\cg{\int_a^b u\,dv} = \cy{\Big[uv\Big]_a^b} - \cb{\int_a^b v\,du}\]

2. Find the product term from the given values. Subtract the product at the lower limit from the product at the upper limit.

   \[\Big[uv\Big]_a^b = u(b)\,v(b) - u(a)\,v(a)\]

3. The new integral \(\int_a^b v\,du\) is often given as a number. If it is an integral of a derivative, the Fundamental Theorem gives its value from the values at the limits.

   \[\int_a^b f'(x)\,dx = f(b) - f(a)\]

4. Subtract the new integral from the product term, which leaves the green region.

   The formula needs \(u'\) and \(v'\) to be continuous on the interval.

When it fails

If the lower limit is 0 and \(u = x\), the product at the lower limit is \(0\cdot v(0) = 0\), so the small rectangle has no area. For other lower limits it does.

Press Move the lower limit away from 0. The small rectangle now has a positive area. Press Leave out the small rectangle. It turns red, because an answer that skips it counts it as part of the green region and is too large.

Example 1: A table of values

The table gives values of \(f\) and \(f'\). Find \(\displaystyle\int_2^5 x f''(x)\,dx\).

1. Take \(u = x\) and \(dv = f''(x)\,dx\). Then \(du = dx\) and \(v = f'(x)\).

   \[\begin{aligned}\int_2^5 x f''(x)\,dx &= \Big[x f'(x)\Big]_2^5 \\ &\quad - \int_2^5 f'(x)\,dx\end{aligned}\]

2. Read \(f'\) from the table. Subtract the product at the lower limit from the product at the upper limit.

   \[\Big[x f'(x)\Big]_2^5 = 5\cdot 1 - 2\cdot 4 = -3\]

3. The new integral is an integral of a derivative, so the Fundamental Theorem gives it from the values of \(f\) at the limits.

   \[\begin{aligned}\int_2^5 f'(x)\,dx &= f(5) - f(2) \\ &= 10 - 3 = 7\end{aligned}\]

4. Subtract the new integral from the product term.

   \[-3 - 7 = -10\]

Example 2: A lower limit that is not zero

The function \(f\) has \(f(1) = 2\) and \(f(4) = 8\). The area under its curve from 1 to 4 is 14. Find \(\displaystyle\int_1^4 x f'(x)\,dx\).

1. Take \(u = x\) and \(dv = f'(x)\,dx\), so \(v = f(x)\). The curve is one graph that fits the facts, and the teal region under it has the given area 14.

   \[\begin{aligned}\int_1^4 x f'(x)\,dx &= \Big[x f(x)\Big]_1^4 \\ &\quad - \int_1^4 f(x)\,dx\end{aligned}\]

2. A tempting wrong move: Take the product term from the upper limit only. Then subtract the given area.

   \[32 - 14 = 18\]

   The green region lies above the height \(f(1) = 2\), so it does not include the red rectangle in the lower left corner. That rectangle has the area \(1\cdot 2 = 2\), and the answer 18 counts it, so it is too large by 2.

3. Evaluate the product at both limits. Subtract the lower product from the upper product.

   \[\Big[x f(x)\Big]_1^4 = 4\cdot 8 - 1\cdot 2 = 30\]

4. Subtract the given area.

   \[30 - 14 = 16\]

Practice

The functions \(f\) and \(g\) have \(f(1) = 2\), \(f(3) = 5\), \(g(1) = 4\) and \(g(3) = 1\). Also \(\displaystyle\int_1^3 f'(x)\,g(x)\,dx = 6\). Find \(\displaystyle\int_1^3 f(x)\,g'(x)\,dx\).

Take \(u = f(x)\) and \(dv = g'(x)\,dx\). The product term needs both functions at both limits, so it uses all four cells of the table.

Questions

The figure shows the graph of a function \(f\) with \(f(0) = 1\) and \(f(5) = 6\). The teal region under the graph has the area \(\tfrac{40}{3} \approx 13.33\).

1. Which expression gives the area of the orange rectangle?

   • \(5\,f(5) - 0\cdot f(0)\)
   • \(f(5) - f(0)\)
   • \(\int_0^5 f(x)\,dx\)
   Answer

   \(5\,f(5) - 0\cdot f(0)\)

2. What is the area of the green region, which is \(\displaystyle\int_0^5 x f'(x)\,dx\)?

   • \(\tfrac{50}{3} \approx 16.67\)
   • \(\tfrac{130}{3} \approx 43.33\)
   • \(\tfrac{40}{3} \approx 13.33\)
   • 30
   Answer

   \(\tfrac{50}{3} \approx 16.67\)

3. Given: \(f(0) = 2\), \(f(2) = 5\) and \(\int_0^2 f(x)\,dx = 7\). What is \(\displaystyle\int_0^2 (x+1)\,f'(x)\,dx\)?

   • 6
   • 8
   • 3
   • 20
   Answer

   6

Summary

You write the integral as \(\int u\,dv\), find the product term from the given values at the two limits, and subtract the new integral.

This works because the green region is the large rectangle, minus the small rectangle, minus the teal region.

Use it when a question gives values of the functions at the limits and one integral, but no formula.

This is one of four skills in topic 6.11 of Which & Why: AP Calculus.