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Amy Ferguson Moncure

Calculus BC · Unit 6

Definite Integrals and Their Properties

Add pieces, reverse limits, and read signed area from a straight graph.

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Key Ideas Easy Medium Hard

\(\int_a^c f=\int_a^b f+\int_b^c f\)

What to Remember

Add and reverse the limits

\(\int_a^c f = \int_a^b f + \int_b^c f\)

\(\int_b^a f = -\int_a^b f\) and \(\int_a^a f = 0\).

Pull out a constant, split a sum

\(\int_a^b kf = k\int_a^b f\)

\(\int_a^b (f+g) = \int_a^b f + \int_a^b g\)

A constant function uses the width

\(\int_a^b c\,dx = c(b-a)\)

Do not replace that integral by the constant alone.

Read signed area from a straight graph

Area above the x-axis is positive. Area below the x-axis is negative. A triangle area is one-half base times height.

Level 1

Easy

Use one property. The given integrals are already evaluated.

Example
Suppose \(\int_0^3 f(x)\,dx = 8\) and \(\int_3^7 f(x)\,dx = -2\). Find \(\int_0^7 f(x)\,dx\).

\(\int_0^7 f = \int_0^3 f + \int_3^7 f\)

\(8 + (-2)\)

\(6\)

Answer: \(6\)

1
Use \(\int_0^4 f(x)\,dx = 5\). Find \(\int_4^0 f(x)\,dx\).
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\(\int_4^0 f = -\int_0^4 f\)

\(-5\)

Answer: \(-5\)

2
\(\int_1^6 g(x)\,dx = 9\). Find \(\int_1^6 3g(x)\,dx\).
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\(\int_1^6 3g = 3\int_1^6 g\)

\(3\cdot 9\)

\(27\)

Answer: \(27\)

3
\(\int_0^2 f(x)\,dx = 4\) and \(\int_0^2 g(x)\,dx = 7\). Find \(\int_0^2 \left(f(x)-g(x)\right)\,dx\).
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\(\int_0^2 (f-g) = \int_0^2 f - \int_0^2 g\)

\(4-7\)

\(-3\)

Answer: \(-3\)

4
Find \(\int_3^3 (x^2+4)\,dx\).
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The two limits are the same.

\(\int_3^3 (x^2+4)\,dx = 0\)

Answer: \(0\)

5
The graph is the segment from \((0, 4)\) to \((4, 0)\).

Graph of f. Segment from (0, 4) to (4, 0).

Find \(\int_0^4 f(x)\,dx\).
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The region is a triangle with base 4 and height 4.

\(\dfrac{1}{2}\cdot 4\cdot 4\)

\(8\)

Answer: \(8\)

Level 2

Medium

Combine two properties, or mix a graph with a constant.

Example
\(\int_0^5 f(x)\,dx = 4\). Find \(\int_0^5 \left(3f(x)+2\right)\,dx\).

\(\int_0^5 (3f+2) = 3\int_0^5 f + \int_0^5 2\,dx\)

\(3\cdot 4\)

\(12\)

\(\int_0^5 2\,dx = 2\cdot 5\)

\(10\)

\(12+10\)

\(22\)

Answer: \(22\)

1
\(\int_0^2 f = 6\), \(\int_2^5 f = -1\), and \(\int_5^8 f = 4\). Find \(\int_0^8 f(x)\,dx\).
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\(\int_0^8 f = 6 + (-1) + 4\)

\(9\)

Answer: \(9\)

2
\(\int_1^4 f(x)\,dx = 10\) and \(\int_1^9 f(x)\,dx = 16\). Find \(\int_4^9 f(x)\,dx\).
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\(\int_1^9 f = \int_1^4 f + \int_4^9 f\)

\(16 = 10 + \int_4^9 f\)

\(6\)

Answer: \(6\)

3
Find \(\int_8^2 4\,dx\).
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\(\int_8^2 4\,dx = -\int_2^8 4\,dx\)

\(\int_2^8 4\,dx = 4\cdot 6\)

\(24\)

\(-24\)

Answer: \(-24\)

4
The graph goes from \((0, 3)\) to \((2, 3)\) to \((6, 0)\).

Graph of f. Horizontal to (2, 3), then down to (6, 0).

Find \(\int_0^6 f(x)\,dx\).
Show Answer

From 0 to 2 the region is a rectangle of height 3.

\(2\cdot 3 = 6\)

From 2 to 6 the region is a triangle with base 4 and height 3.

\(\dfrac{1}{2}\cdot 4\cdot 3 = 6\)

\(6+6\)

\(12\)

Answer: \(12\)

5
A semicircle of radius 3 sits above the x-axis on \([-3, 3]\). Find \(\int_{-3}^{3} f(x)\,dx\).
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Area of a full circle of radius 3 is \(\pi\cdot 3^2\).

\(9\pi\)

A semicircle is half of that.

\(\dfrac{9\pi}{2}\)

Answer: \(\dfrac{9\pi}{2}\)

Level 3

Hard

Signed area, a missing piece, or a property used in the wrong order.

Example
The graph is the segment from \((0, 0)\) to \((4, -4)\).

Graph of f. Segment from (0, 0) to (4, −4).

Find \(\int_0^4 f(x)\,dx\).

The triangle has base 4 and height 4, and it lies below the x-axis.

\(\dfrac{1}{2}\cdot 4\cdot 4\)

\(8\)

Signed area is the opposite of that area.

\(-8\)

Answer: \(-8\)

1
\(\int_0^6 f(x)\,dx = 11\) and \(\int_0^6 g(x)\,dx = -4\). Find \(\int_0^6 \left(2f(x)-3g(x)\right)\,dx\).
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\(2\int_0^6 f - 3\int_0^6 g\)

\(2\cdot 11 - 3(-4)\)

\(22 - (-12)\)

\(22+12\)

\(34\)

Answer: \(34\)

2
\(\int_2^2 f(x)\,dx = 0\), \(\int_2^5 f(x)\,dx = 7\), and \(\int_5^1 f(x)\,dx = -3\). Find \(\int_1^2 f(x)\,dx\).
Show Answer

Going from 1 to 5 can pass through 2.

\(\int_1^5 f = \int_1^2 f + \int_2^5 f\)

Also \(\int_1^5 f = -\int_5^1 f = 3\).

\(3 = \int_1^2 f + 7\)

\(\int_1^2 f = 3-7\)

\(-4\)

Answer: \(-4\)

3
The graph of \(f\) on \([0, 4]\) is a semicircle of radius 2 below the x-axis. Find \(\int_0^4 f(x)\,dx\).
Show Answer

A full circle of radius 2 has area \(4\pi\).

The semicircle has area \(2\pi\).

It lies below the x-axis, so the integral is negative.

\(-2\pi\)

Answer: \(-2\pi\)

4
Which statement must be true?
A. \(\int_0^4 (f(x)+g(x))\,dx = \int_0^4 f(x)\,dx + \int_0^4 g(x)\,dx\)
B. \(\int_0^4 f(x)g(x)\,dx = \left(\int_0^4 f(x)\,dx\right)\left(\int_0^4 g(x)\,dx\right)\)
C. \(\int_0^4 f(x)\,dx = f(4)-f(0)\)
D. \(\int_4^0 f(x)\,dx = \int_0^4 f(x)\,dx\)
Show Answer

Integrals add when the integrand is a sum.

A product of functions is not the product of the integrals.

\(f(4)-f(0)\) is the integral of \(f'\), not of \(f\).

Reversing the limits changes the sign.

Answer: A. \(\int_0^4 (f+g)=\int_0^4 f+\int_0^4 g\)

5
\(\int_{-2}^{2} f(x)\,dx = 5\) and \(f\) is odd. Find \(\int_0^{2} f(x)\,dx\).
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An odd function has opposite signed areas on \([-2, 0]\) and \([0, 2]\).

Those two areas add to 0 when the limits are symmetric.

The given integral is 5, not 0, so \(f\) cannot be odd on this interval.

No value is determined. The given facts contradict each other.

Answer: impossible for an odd function

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