Calculus 1

Properties of definite integrals

Once a few integrals are known, how can others be found from them without going back to Riemann sums, and how much can be said about an integral that has no formula at all?

You need

Even and odd functionsAlgebra 2

Unlocks

Improper integralsCalculus 2

Where you are 3 in The definite integral

Calculus8

Integrals16

The definite integral3

Properties of definite integrals

Signed area and integrals from geometry The definite integral as a limit of Riemann sums

Why this exists

The definite integral as a limit of Riemann sums defines \int_a^b f(x)\,dx as the limit of Riemann sums, and even \int_1^3 x^2\,dx takes a page that way. Nobody computes every integral from scratch. Integrals are assembled from ones already known, using rules that every Riemann sum obeys and that therefore pass to the limit: the integral of a sum is the sum of the integrals, and an integral over a long interval is the sum of the integrals over its parts.

Some integrals, such as \displaystyle \int_0^1 \sqrt{1 + x^4}\,dx, have no formula in the usual functions at all. Comparison rules still trap such a number between two values that can be computed, and say how large an integral can be from bounds on its integrand.

Definition

Day 1: the integral for every pair of limits

For a < b and f integrable on [a, b], the definite integral is the limit of Riemann sums (The definite integral as a limit of Riemann sums). It is extended to every pair of limits by two conventions, met first in Signed area and integrals from geometry:

\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx,

\int_a^a f(x)\,dx = 0.

The properties of the definite integral are the theorems below about this extended integral. Each one is a property of finite sums carried through the limit.

Day 2: symmetry and bounds

A function is even when f(-x) = f(x) for every x in its domain and odd when f(-x) = -f(x) (Even and odd functions). Powers x^2, x^4 and \cos x are even; x, x^3 and \sin x are odd. A bound for f on [a, b] is a pair of numbers m and M with m \le f(x) \le M for every x in [a, b].

Intuition

Day 1: cutting and stacking

In Figure 1 a dashed line at b cuts the region over [a, c] into two pieces, so the areas add. The rule survives when b lies outside [a, c]. If a < c < b, the integral from a to b overshoots c, and the integral from b back to c, which is negative by the reversal convention, takes the overshoot away.

Figure 1Cutting the interval at b cuts the region, and the two integrals add to \int_a^c f(x)\,dx.

Multiplying f by 3 makes every rectangle in every Riemann sum three times as tall, so every sum, and the limit, triples. Adding g to f stacks each rectangle for g on the matching rectangle for f, so the sums add.

Day 2: pairing and squeezing

For an odd function the heights at x and -x are opposite. In Figure 2 the midpoints of the eight pieces come in mirror pairs, and each pair of rectangles cancels exactly, so every midpoint sum is 0, not merely its limit. Signed area and integrals from geometry showed the same thing with a half turn of the region; here the cancellation happens inside each sum, which is what makes it a proof from the definition.

Figure 2For an odd function the midpoints come in pairs \pm m, and each pair of terms f(m)\,\Delta x + f(-m)\,\Delta x is 0.

If the graph of f never rises above the graph of g, each rectangle for f is no taller than the matching one for g, so the integral of f is no larger. A graph caught between heights m and M has its integral caught between two rectangles.

Results

Constant multiple and sum (linearity). If f and g are integrable on [a, b] and c is a constant, then

\begin{aligned} \int_a^b c\,f(x)\,dx &= c\int_a^b f(x)\,dx, \\ \int_a^b \big(f(x) \pm g(x)\big)\,dx &= \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx. \end{aligned}

Additivity. If f is integrable on an interval containing a, b and c, in any order, then

\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx.

Three integrals from the definition. For all real a, b and every constant k,

\begin{aligned} \int_a^b k\,dx &= k(b - a), \qquad \int_a^b x\,dx = \frac{b^2 - a^2}{2}, \\ \int_a^b x^2\,dx &= \frac{b^3 - a^3}{3}. \end{aligned}

Symmetric intervals. Let f be integrable on [-a, a]. If f is odd, \int_{-a}^{a} f(x)\,dx = 0. If f is even, \int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx.

Comparison. Let a < b and let f and g be integrable on [a, b].

If f(x) \ge 0 on [a, b], then \int_a^b f(x)\,dx \ge 0.
If f(x) \le g(x) on [a, b], then \int_a^b f(x)\,dx \le \int_a^b g(x)\,dx.
If m \le f(x) \le M on [a, b], then m(b - a) \le \int_a^b f(x)\,dx \le M(b - a).

Absolute value. If f is continuous on [a, b] with a < b, then \left|\int_a^b f(x)\,dx\right| \le \int_a^b |f(x)|\,dx.

Why it is true

Linearity. Take one partition of [a, b] and one choice of sample points. A finite sum may be regrouped and reordered and a common factor pulled out, so the Riemann sum for c\,f + g is

\begin{aligned} &\sum \big(c\,f(x_i^*) + g(x_i^*)\big)\Delta x_i \\ &\quad = c\sum f(x_i^*)\,\Delta x_i + \sum g(x_i^*)\,\Delta x_i. \end{aligned}

As the mesh goes to 0 the two sums on the right approach \int_a^b f(x)\,dx and \int_a^b g(x)\,dx, so by the limit laws for sums and constant multiples the left side approaches c\int_a^b f(x)\,dx + \int_a^b g(x)\,dx. Taking g = 0 gives the constant multiple rule, and c = 1 gives the sum rule; the difference rule follows from these two, since f - g = f + (-1)g (Exercise 15). With the limits reversed, both sides change sign, so the rules hold for every pair of limits.

Additivity when a < b < c. A function integrable on [a, c] is integrable on [a, b] and on [b, c] (stated without proof). Take partitions of [a, b] and of [b, c] with meshes going to 0. Together they form a partition of [a, c] with b as one of its points, and its Riemann sum is the sum over [a, b] plus the sum over [b, c]. The left side approaches \int_a^c f(x)\,dx, because integrability allows every partition, and the two parts approach \int_a^b f(x)\,dx and \int_b^c f(x)\,dx.

Every other order. Suppose a < c < b. The case just proved, applied to a < c < b, gives \int_a^c f + \int_c^b f = \int_a^b f. Move \int_c^b f across and use the reversal convention:

\begin{aligned} \int_a^c f(x)\,dx &= \int_a^b f(x)\,dx - \int_c^b f(x)\,dx \\ &= \int_a^b f(x)\,dx + \int_b^c f(x)\,dx. \end{aligned}

The other four orders of three different points go the same way. When two of the points are equal, additivity reduces at once to one of the two conventions. The conventions are exactly what this needs, and no other choice works. Putting a = b = c in additivity gives \int_a^a f = 2\int_a^a f, so \int_a^a f = 0. Putting c = a then gives \int_a^b f + \int_b^a f = \int_a^a f = 0, which is the reversal rule.

The three integrals. The first is the constant rule of The definite integral as a limit of Riemann sums. For b > 0, right endpoints with \displaystyle \Delta x = \frac{b}{n} and \displaystyle x_i = \frac{bi}{n} give

\begin{aligned} \sum_{i=1}^{n} \frac{b^2 i^2}{n^2}\cdot\frac{b}{n} &= \frac{b^3}{n^3}\cdot\frac{n(n+1)(2n+1)}{6} \\ &\to \frac{b^3}{3}, \end{aligned}

and the same computation with \displaystyle \sum i = \frac{n(n+1)}{2} gives \displaystyle \int_0^b x\,dx = \frac{b^2}{2}. For b < 0 the same points \displaystyle x_i = \frac{bi}{n} cut [b, 0] into n pieces of width \displaystyle -\frac{b}{n}, and the same sums, with that positive width, give \displaystyle \int_b^0 x^2\,dx = -\frac{b^3}{3} and \displaystyle \int_b^0 x\,dx = -\frac{b^2}{2}; the reversal convention turns these into \displaystyle \frac{b^3}{3} and \displaystyle \frac{b^2}{2}, so both formulas hold for every b. Additivity then gives \int_a^b = \int_a^0 + \int_0^b = \int_0^b - \int_0^a, which is \displaystyle \frac{b^3 - a^3}{3} and \displaystyle \frac{b^2 - a^2}{2}.

Odd and even functions, from the sums. Cut [-a, a] into 2n equal pieces of width \displaystyle \Delta x = \frac{a}{n}. The grid points -a, -a + \Delta x, \dots, a are carried to themselves by x \mapsto -x, so the midpoints of the n left pieces are exactly the negatives -m_1, \dots, -m_n of the midpoints m_1, \dots, m_n of the n right pieces. The midpoint sum is

M_{2n} = \sum_{i=1}^{n} \big(f(m_i) + f(-m_i)\big)\Delta x.

If f is odd, each bracket is f(m_i) - f(m_i) = 0, so M_{2n} = 0 for every n and its limit, the integral, is 0. If f is even, each bracket is 2f(m_i), so M_{2n} is twice the midpoint sum for \int_0^a f(x)\,dx with n pieces, and the limits give 2\int_0^a f(x)\,dx. Integrability is what allows the limit to be taken along midpoint sums alone.

Comparison. If f \ge 0 and a < b, every term f(x_i^*)\,\Delta x_i is at least 0, so every Riemann sum is at least 0. A limit of numbers that are all at least 0 is at least 0: if the limit L were negative, the sums would eventually lie within \displaystyle \frac{|L|}{2} of L and be negative. For f \le g, apply this to g - f \ge 0 and use linearity. For m \le f \le M, compare f with the constants m and M, whose integrals are m(b - a) and M(b - a).

Absolute value. For continuous f, |f| is continuous and so integrable. Since -|f(x)| \le f(x) \le |f(x)| for every x, comparison gives

\begin{aligned} -\int_a^b |f(x)|\,dx &\le \int_a^b f(x)\,dx \\ &\le \int_a^b |f(x)|\,dx, \end{aligned}

which says \left|\int_a^b f(x)\,dx\right| \le \int_a^b |f(x)|\,dx.

Worked examples

Example 1

Day 1: Ifan combines two known integrals

Ifan knows \int_2^6 f(x)\,dx = 5 and \int_2^6 g(x)\,dx = -3. Find (a) \int_2^6 \big(4f(x) - 2g(x) + 3\big)\,dx and (b) \int_6^2 \big(f(x) + g(x)\big)\,dx.

Solution

(a) By the sum rule the integral splits into three, and by the constant multiple rule the constants come out:

\begin{aligned} &4\int_2^6 f(x)\,dx - 2\int_2^6 g(x)\,dx + \int_2^6 3\,dx \\ &\quad = 4(5) - 2(-3) + 3(6 - 2) = 38. \end{aligned}

The constant 3 is a function too: its integral is 3 times the length of [2, 6], not 3.

(b) Reversing the limits changes the sign, and then the sum rule applies: \int_6^2 (f + g)\,dx = -\left(\int_2^6 f\,dx + \int_2^6 g\,dx\right) = -(5 - 3) = -2.

Example 2

Day 1: Cadi’s table of integrals

Cadi is given \int_0^4 f(x)\,dx = 10, \int_0^9 f(x)\,dx = 3 and \int_6^9 f(x)\,dx = -2. Find (a) \int_4^9 f(x)\,dx, (b) \int_4^6 f(x)\,dx and (c) \int_6^0 f(x)\,dx.

Solution

(a) Additivity with 0 < 4 < 9 gives \int_0^4 f + \int_4^9 f = \int_0^9 f, so \int_4^9 f(x)\,dx = 3 - 10 = -7.

(b) Additivity holds in any order. Going from 4 to 9 and then back to 6:

\begin{aligned} \int_4^6 f(x)\,dx &= \int_4^9 f(x)\,dx + \int_9^6 f(x)\,dx \\ &= -7 - (-2) = -5. \end{aligned}

(c) First \int_0^6 f = \int_0^4 f + \int_4^6 f = 10 - 5 = 5; reversing, \int_6^0 f(x)\,dx = -5. As a check, \int_0^6 f + \int_6^9 f = 5 - 2 = 3, which matches \int_0^9 f.

Example 3

Day 1: Osian integrates a quadratic without sums

Find \int_1^3 \left(3x^2 - 4x + 5\right)dx.

Solution

Linearity splits the integrand into its terms, and each term is one of the three integrals in the Results:

\begin{aligned} &\int_1^3 \left(3x^2 - 4x + 5\right)dx \\ &\quad = 3\cdot\frac{3^3 - 1^3}{3} - 4\cdot\frac{3^2 - 1^2}{2} + 5(3 - 1) \\ &\quad = 26 - 16 + 10 = 20. \end{aligned}

From the definition this would mean expanding 3x_i^2 - 4x_i + 5 with \displaystyle x_i = 1 + \frac{2i}{n} and taking a limit of three sum formulas. Linearity does that work once, for x^2 and x, and reuses it.

Example 4

Day 1: a piecewise integrand

Let f(x) = x^2 for x \le 2 and f(x) = 8 - 2x for x > 2. Find \int_0^5 f(x)\,dx.

Solution

The two formulas agree at x = 2, where both give 4, so f is continuous and integrable on [0, 5]. Additivity cuts the interval where the formula changes:

\int_0^5 f(x)\,dx = \int_0^2 x^2\,dx + \int_2^5 (8 - 2x)\,dx.

The first piece is \displaystyle \frac{2^3 - 0}{3} = \frac{8}{3}. For the second, the line y = 8 - 2x falls from 4 at x = 2 through 0 at x = 4 to -2 at x = 5: a triangle of area \displaystyle \frac{1}{2}\cdot 2\cdot 4 = 4 above the axis and one of area \displaystyle \frac{1}{2}\cdot 1\cdot 2 = 1 below, so it is 4 - 1 = 3 (Signed area and integrals from geometry). Altogether \displaystyle \int_0^5 f(x)\,dx = \frac{8}{3} + 3 = \frac{17}{3}.

Example 5

Day 2: odd parts vanish

Evaluate (a) \int_{-2}^{2} \left(x^5 - 3x^3 + x^2 + 4\right)dx and (b) \displaystyle \int_{-2}^{2} \frac{x^3}{1 + x^2}\,dx.

Solution

(a) Linearity splits the integrand into x^5 - 3x^3, which is odd because every power is odd, and x^2 + 4, which is even. Both parts are continuous. The odd part integrates to 0 over [-2, 2], and the even part is twice its integral over [0, 2]:

\begin{aligned} \int_{-2}^{2} \left(x^2 + 4\right)dx &= 2\left(\frac{2^3}{3} + 4\cdot 2\right) \\ &= \frac{64}{3}. \end{aligned}

(b) With \displaystyle f(x) = \frac{x^3}{1 + x^2}, \displaystyle f(-x) = \frac{-x^3}{1 + x^2} = -f(x), and the denominator is never 0, so f is odd and continuous on [-2, 2]. The integral is 0. No formula for an antiderivative was needed, and none from this unit’s tools would have found the value.

Example 6

Day 2: Cadi traps an integral with no formula

The integral \displaystyle I = \int_0^1 \sqrt{1 + x^4}\,dx has no antiderivative among the usual functions. Find bounds for it, first from the largest and smallest values of the integrand, then from a comparison function.

Solution

On [0, 1], x^4 runs from 0 to 1, so \displaystyle 1 \le \sqrt{1 + x^4} \le \sqrt{2}. With m = 1, \displaystyle M = \sqrt{2} and b - a = 1:

1 \le I \le \sqrt{2} \approx 1.414.

For a tighter upper bound, compare with \displaystyle 1 + \frac{x^2}{2} (Figure 3). Both sides of \displaystyle \sqrt{1 + x^4} \le 1 + \frac{x^4}{2} are positive, and squaring the right gives \displaystyle 1 + x^4 + \frac{x^8}{4} \ge 1 + x^4, so it holds. On [0, 1], x^4 \le x^2, so \displaystyle \sqrt{1 + x^4} \le 1 + \frac{x^2}{2}. Comparison and the three integrals give

\begin{aligned} 1 \le I &\le \int_0^1 \left(1 + \frac{x^2}{2}\right)dx \\ &= 1 + \frac{1}{6} = \frac{7}{6} \approx 1.167. \end{aligned}

Figure 3Example 6. The graph lies between y = 1 and y = 1 + \tfrac{x^2}{2}, which traps the shaded area more tightly than the box of height \displaystyle \sqrt{2}.

A computer adding midpoint sums gives I \approx 1.089, inside both intervals.

Example 7

Day 2: Osian bounds an oscillating integral

Show that \displaystyle \left|\int_0^2 x^2 \sin(5x)\,dx\right| \le \frac{8}{3}.

Solution

\begin{aligned} \left|\int_0^2 x^2 \sin(5x)\,dx\right| &\le \int_0^2 x^2\left|\sin(5x)\right|dx \\ &\le \int_0^2 x^2\,dx = \frac{8}{3}. \end{aligned}

The true value is about 0.555, far inside the bound: the positive and negative stretches mostly cancel, which the absolute value ignores. The bound is what can be said with no computing at all, and it holds for every integrand x^2 \sin(kx) on [0, 2].

Boundaries

Every property needs the integrals in it to exist. Additivity needs f integrable on the whole interval from the smallest to the largest of a, b, c: \displaystyle \frac{1}{x^2} has integrals over [1, 2] and [-2, -1], but none over [-1, 1], so nothing over [-1, 1] can be built from them.
There is no product rule: \int_a^b f(x)g(x)\,dx is in general not \int_a^b f\,dx \cdot \int_a^b g\,dx (Exercise 14), and there is no quotient rule either.
The symmetry rules need an interval symmetric about 0 and an integrand integrable on it. Over [0, a] or [-a, b] with b \ne a an odd function need not integrate to 0.
The comparison rules need a < b. With the limits reversed every inequality flips (Exercise 24).
Comparison is not strict at single points: changing f at one point does not change its integral, so f < g at a few points says nothing more than f \le g. For continuous f \le g with f(x) < g(x) somewhere, the inequality of integrals is strict (stated without proof).
For continuous f, \left|\int_a^b f\right| = \int_a^b |f| exactly when f does not change sign on [a, b].

Where this goes next

The fundamental theorem of calculus replaces the short table of three integrals with an antiderivative for every elementary function, and The second fundamental theorem of calculus uses linearity and additivity at every step, especially for piecewise and absolute value integrands. Additivity is also what makes the accumulation function \int_a^x f(t)\,dt behave: moving x adds or removes a thin strip. Dividing m(b - a) \le \int_a^b f \le M(b - a) by b - a previews the average value of a function. Comparison returns in Improper integrals, where an integral out to infinity is shown to be finite by trapping it under a larger one.

Exercises

Day 1: combining known integrals

Use \int_1^5 f(x)\,dx = 6 and \int_1^5 g(x)\,dx = -2 in Exercises 1 to 4.

Find \int_1^5 \big(3f(x) - g(x)\big)\,dx.
Find \int_1^5 \big(2g(x) - 7\big)\,dx.
Find \int_5^1 \big(f(x) + 4g(x)\big)\,dx.
Find the constant k with \int_1^5 \big(f(x) + k\big)\,dx = 0.

Day 1: splitting and joining intervals

Use \int_{-2}^{3} h(x)\,dx = 5, \int_1^3 h(x)\,dx = 9 and \int_1^7 h(x)\,dx = -4 in Exercises 5 to 8.

Find \int_{-2}^{1} h(x)\,dx.
Find \int_3^7 h(x)\,dx.
Find \int_{-2}^{7} h(x)\,dx.
Find \int_7^{-2} \big(2h(x) + 1\big)\,dx.

Day 1: polynomials and piecewise integrands

Use the three integrals in the Results.

Evaluate \int_{-1}^{2} \left(x^2 - 2x + 3\right)dx.
Evaluate \int_2^4 \left(6x^2 - 5\right)dx.
Evaluate \int_1^4 (x - 1)(x - 3)\,dx, and explain the value from the graph.
Let f(x) = 4 for 0 \le x < 2 and f(x) = x - 5 for 2 \le x \le 7. Evaluate \int_0^7 f(x)\,dx.
Evaluate \int_{-1}^{3} \left|x^2 - 1\right|dx.

Day 1: reasoning

Cadi guesses that \int_0^2 f(x)g(x)\,dx = \int_0^2 f(x)\,dx \cdot \int_0^2 g(x)\,dx. Show with f(x) = g(x) = x that this is false.
Using only the sum rule and the constant multiple rule, prove \int_a^b \big(f(x) - g(x)\big)\,dx = \int_a^b f(x)\,dx - \int_a^b g(x)\,dx.

Day 2: symmetry

Evaluate \int_{-3}^{3} \left(|x| + x^5\right)dx.
Evaluate \displaystyle \int_{-1}^{1} \frac{\sin x}{1 + x^2}\,dx.
Let f be continuous and even with \int_{-4}^{4} f(x)\,dx = 10 and \int_0^1 f(x)\,dx = 2. Find \int_1^4 f(x)\,dx and \int_{-4}^{-1} f(x)\,dx.
Evaluate \int_{-2}^{2} (x + 1)^2\,dx by splitting the integrand into an even part and an odd part.

Day 2: comparison and bounds

Show that \displaystyle \frac{2}{9} \le \int_0^2 \frac{1}{1 + x^3}\,dx \le 2.
Using e^t \ge 1 + t for every t, show that \displaystyle \frac{4}{3} \le \int_0^1 e^{x^2}\,dx \le e.
Without computing either integral, decide which is larger: (a) \displaystyle \int_0^1 \sqrt{x}\,dx or \int_0^1 x\,dx; (b) \displaystyle \int_1^2 \sqrt{x}\,dx or \int_1^2 x\,dx.
Show that \displaystyle \left|\int_0^3 x \sin(x^4)\,dx\right| \le \frac{9}{2}.

Day 2: reasoning

Ifan says that f(x) \le g(x) on the interval always gives \int_a^b f(x)\,dx \le \int_a^b g(x)\,dx. Give an example with b < a where it fails, and explain why the proof needs a < b.
Let f be continuous and odd, and 0 < a < b. Prove that \int_{-a}^{b} f(x)\,dx = \int_a^b f(x)\,dx.

Answers

20. 3(6) - (-2) = 20.
-32. 2(-2) - 7(5 - 1) = -4 - 28.
2. \int_5^1 (f + 4g)\,dx = -\big(6 + 4(-2)\big) = 2.
\displaystyle k = -\frac{3}{2}. 6 + k(5 - 1) = 0.
-4. \int_{-2}^{1} h + \int_1^3 h = \int_{-2}^{3} h, so 5 - 9.
-13. \int_1^3 h + \int_3^7 h = \int_1^7 h, so -4 - 9.
-8. \int_{-2}^{1} h + \int_1^7 h = -4 + (-4). Check: \int_{-2}^{3} h + \int_3^7 h = 5 - 13 = -8.
7. \int_{-2}^{7} (2h + 1)\,dx = 2(-8) + 9 = -7, and reversing the limits gives 7.
9. \displaystyle \frac{8 - (-1)}{3} - 2\cdot\frac{4 - 1}{2} + 3\cdot 3 = 3 - 3 + 9.
102. \displaystyle 6\cdot\frac{64 - 8}{3} - 5\cdot 2 = 112 - 10.
0. Expanding, x^2 - 4x + 3: \displaystyle \frac{64 - 1}{3} - 4\cdot\frac{16 - 1}{2} + 3\cdot 3 = 21 - 30 + 9 = 0. The parabola is below the axis on (1, 3) and above it on (3, 4), and the two areas happen to be equal, each \displaystyle \frac{4}{3}.
\displaystyle \frac{11}{2}. \int_0^2 4\,dx = 8. On [2, 7] the line runs from -3 to 2, crossing at 5: triangles \displaystyle -\frac{9}{2} and +2, total \displaystyle -\frac{5}{2}. The jump at 2 does not matter: f is bounded with one jump, so it is integrable, and additivity cuts there. \displaystyle 8 - \frac{5}{2} = \frac{11}{2}.
8. |x^2 - 1| = 1 - x^2 on [-1, 1] and x^2 - 1 on [1, 3]. The pieces are \displaystyle 2 - \frac{2}{3} = \frac{4}{3} and \displaystyle \frac{26}{3} - 2 = \frac{20}{3}, total \displaystyle \frac{24}{3} = 8.
\displaystyle \int_0^2 x^2\,dx = \frac{8}{3}, but \int_0^2 x\,dx \cdot \int_0^2 x\,dx = 2 \cdot 2 = 4.
f - g = f + (-1)g. The sum rule gives \int (f + (-1)g) = \int f + \int (-1)g, and the constant multiple rule gives \int (-1)g = -\int g.
9. x^5 is odd and contributes 0; |x| is even, so \displaystyle 2\int_0^3 x\,dx = 2\cdot\frac{9}{2} = 9.
0. \displaystyle \frac{\sin(-x)}{1 + (-x)^2} = -\frac{\sin x}{1 + x^2}, and the integrand is continuous, so it is odd on [-1, 1].
Both are 3. Even gives \int_0^4 f = 5, so \int_1^4 f = 5 - 2 = 3. Reflection gives \int_{-4}^{-1} f = \int_1^4 f; or \int_{-4}^{-1} f = \int_{-4}^{0} f - \int_{-1}^{0} f = 5 - 2.
\displaystyle \frac{28}{3}. (x + 1)^2 = (x^2 + 1) + 2x; the odd part 2x gives 0, and \displaystyle 2\int_0^2 (x^2 + 1)\,dx = 2\left(\frac{8}{3} + 2\right).
On [0, 2], x^3 runs from 0 to 8, so \displaystyle \frac{1}{9} \le \frac{1}{1 + x^3} \le 1. Multiply the bounds by the length 2.
Lower: e^{x^2} \ge 1 + x^2, and \displaystyle \int_0^1 (1 + x^2)\,dx = 1 + \frac{1}{3} = \frac{4}{3}. Upper: x^2 \le 1 on [0, 1], so e^{x^2} \le e, and e \cdot 1 = e. (The value is about 1.463.)
(a) \displaystyle \int_0^1 \sqrt{x}\,dx: for 0 \le x \le 1, \displaystyle x = \sqrt{x}\cdot\sqrt{x} \le \sqrt{x}. (b) \int_1^2 x\,dx: for x \ge 1, \displaystyle \sqrt{x} \le \sqrt{x}\cdot\sqrt{x} = x. In both cases the functions differ on more than a point and are continuous, so the larger integral is strictly larger.
The integrand is continuous and |x\sin(x^4)| \le x on [0, 3], so \displaystyle \left|\int_0^3 x\sin(x^4)\,dx\right| \le \int_0^3 x\,dx = \frac{9}{2}.
f = 0, g = 1, a = 1, b = 0: \int_1^0 0\,dx = 0 but \int_1^0 1\,dx = -1, and 0 > -1. The proof uses \Delta x_i > 0 to make each term \big(g(x_i^*) - f(x_i^*)\big)\Delta x_i at least 0; with b < a every \Delta x_i is negative.
Additivity with -a < a < b gives \int_{-a}^{b} f = \int_{-a}^{a} f + \int_a^b f, and the first integral is 0 because f is odd and continuous on [-a, a].