IB Mathematics: Analysis and Approaches Quiz: Numerical Integration

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Numerical IntegrationQuestion 1 of 20

Let $I = \int_0^4 f(x) dx$ . The trapezoidal rule with $n=4$ gives an approximation $T_4=12$ . A new function is defined as $g(x) = 3f(x) - 1$ . What is the trapezoidal rule approximation for $\int_0^4 g(x) dx$ with $n=4$ ?

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Numerical Integration

Practice Numerical Integration in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Numerical Integration, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

32 (correct answer)
35
36
38

Explanation: We use the linearity property of the trapezoidal rule approximation. Let

T_n(f)

denote the approximation for a function

f

.\n

T_n(ag+bh) = aT_n(g) + bT_n(h)

for constants

a, b

.\nWe want to find

T_4

for

\int_0^4 g(x) dx = \int_0^4 (3f(x) - 1) dx

.\nUsing the linearity property:\n

T_4(g) = T_4(3f - 1) = 3T_4(f) - T_4(1)

.\nWe are given

T_4(f) = 12

.\nWe need to calculate

T_4(1)

, which is the trapezoidal approximation for

\int_0^4 1 dx

. Since the function

h(x)=1

is linear, the trapezoidal rule is exact.\n

\int_0^4 1 dx = [x]_0^4 = 4

.\nSo,

T_4(1) = 4

.\nSubstituting these values back:\n

T_4(g) = 3(12) - 4 = 36 - 4 = 32

.\nA) This is the correct value.\nB) This results from

3(12)-1=35

, incorrectly subtracting the constant 1 instead of its integral approximation.\nC) This results from

3(12)=36

, ignoring the constant term -1 entirely.\nD) This results from

3(12)+2=38

, possibly making a sign error and a calculation error on the constant part.

Question 2

Let $T_1$ and $T_2$ be the trapezoidal approximations for $\int_0^2 f(x) dx$ with $n=1$ and $n=2$ subintervals, respectively. Which of the following correctly expresses $T_2$ in terms of $T_1$ and a function value?

$T_2 = \frac{1}{2}T_1 + 2f(1)$
$T_2 = \frac{1}{2}(T_1 + f(1))$
$T_2 = T_1 + f(1)$
$T_2 = \frac{1}{2}T_1 + f(1)$ (correct answer)

Explanation: First, let's write out the expressions for

T_1

and

T_2

.\nFor

T_1

, we have

n=1

[0, 2]

, so

h=2

. The points are

x_0=0, x_1=2

.\n

T_1 = \frac{h}{2}[f(0) + f(2)] = \frac{2}{2}[f(0) + f(2)] = f(0) + f(2)

.\nFor

T_2

, we have

n=2

[0, 2]

, so

h=1

. The points are

x_0=0, x_1=1, x_2=2

.\n

T_2 = \frac{h}{2}[f(0) + 2f(1) + f(2)] = \frac{1}{2}[f(0) + 2f(1) + f(2)]

.\nNow we substitute the expression for

T_1

into the expression for

T_2

. We can rewrite the

T_2

expression as:\n

T_2 = \frac{1}{2}[(f(0) + f(2)) + 2f(1)]

.\nSince

T_1 = f(0) + f(2)

, we can substitute it in:\n

T_2 = \frac{1}{2}[T_1 + 2f(1)] = \frac{1}{2}T_1 + f(1)

.\nDistractors A, B, and C are incorrect algebraic manipulations of the relationship.

Question 3

A table of values for a function $g(x)$ is given below.

x	2	3	4	5	6
g(x)	4	7	11	16	22

Use the trapezoidal rule with 4 subintervals to estimate $\int_0^4 g(x+2) dx$ .

30
47 (correct answer)
52
94

Explanation: First, we perform a substitution to simplify the integral. Let

u = x+2

. Then

du = dx

. We must also change the limits of integration: when

x=0

u=2

; when

x=4

u=6

. The integral becomes

\int_2^6 g(u) du

.\nNow we use the trapezoidal rule to estimate this new integral with

n=4

subintervals, using the values from the table. The interval is

[2, 6]

, so the width of each subinterval is

h = \frac{6-2}{4} = 1

. The u-values are 2, 3, 4, 5, 6, which match the table.\n

T_4 = \frac{h}{2}[g(2) + 2g(3) + 2g(4) + 2g(5) + g(6)]

T_4 = \frac{1}{2}[4 + 2(7) + 2(11) + 2(16) + 22]

T_4 = \frac{1}{2}[4 + 14 + 22 + 32 + 22]

T_4 = \frac{1}{2}[94] = 47

.\nA) This results from forgetting to multiply the middle terms by 2:

\frac{1}{2}(4+7+11+16+22) = 30

.\nC) This is the approximation with only one interval (n=1):

\frac{6-2}{2}(g(2)+g(6)) = 2(4+22) = 52

.\nD) This is the sum inside the brackets, forgetting the factor of

\frac{1}{2}

Question 4

A function $f(x)$ is decreasing and concave down on the interval $[a, b]$ . Let $I = \int_a^b f(x) dx$ , and let $L_n$ , $R_n$ , and $T_n$ be the left-hand Riemann sum, right-hand Riemann sum, and trapezoidal rule approximations with $n$ subintervals, respectively. Which of the following inequalities is always true for $n \ge 1$ ?

$R_n < T_n < I < L_n$ (correct answer)
$L_n < I < T_n < R_n$
$T_n < R_n < I < L_n$
$R_n < I < L_n < T_n$

Explanation: Let's analyze the properties of the approximations based on the function's characteristics.\n1. Decreasing function: For a decreasing function, the left endpoint of any subinterval is the maximum value and the right endpoint is the minimum. Therefore, the left-hand Riemann sum (

L_n

) will be an overestimate, and the right-hand Riemann sum (

R_n

) will be an underestimate. This gives us the inequality

R_n < I < L_n

.\n2. Concave down function: For a function that is concave down, the line segment connecting two points on the curve (the top of a trapezoid) lies below the curve. Therefore, the area of the trapezoid will be less than the area under the curve. This means the trapezoidal rule (

T_n

) provides an underestimate, so

T_n < I

.\n3. Comparing $R_n$ and $T_n$ : The formula for the trapezoidal rule is the average of the left and right Riemann sums:

T_n = \frac{L_n + R_n}{2}

. Since

R_n < L_n

, it follows that

R_n < \frac{L_n + R_n}{2} < L_n

, which means

R_n < T_n < L_n

.\nCombining these facts: We know

R_n < T_n

and

T_n < I

and

I < L_n

. Putting them all together gives the complete ordering:

R_n < T_n < I < L_n

Question 5

The trapezoidal rule is used to estimate $\int_a^b f(x) dx$ . For which of the following functions and intervals is the estimate guaranteed to be an overestimate?

$f(x) = \ln(x)$ on $[1, 2]$
$f(x) = e^{-x}$ on $[0, 1]$ (correct answer)
$f(x) = \cos(x)$ on $[-\frac{\pi}{2}, \frac{\pi}{2}]$
$f(x) = x^3$ on $[-1, 1]$

Explanation: The trapezoidal rule provides an overestimate when the function is concave up, which means its second derivative is positive (

f''(x) > 0

) on the interval of integration. We check the second derivative for each option.\nA) For

f(x) = \ln(x)

f'(x) = 1/x

and

f''(x) = -1/x^2

. On

[1, 2]

f''(x) < 0

, so the function is concave down, leading to an underestimate.\nB) For

f(x) = e^{-x}

f'(x) = -e^{-x}

and

f''(x) = e^{-x}

. On

[0, 1]

e^{-x}

is always positive, so the function is concave up, leading to an overestimate.\nC) For

f(x) = \cos(x)

f''(x) = -\cos(x)

. On

[-\frac{\pi}{2}, \frac{\pi}{2}]

\cos(x) \ge 0

, so

f''(x) \le 0

. The function is concave down, leading to an underestimate.\nD) For

f(x) = x^3

f''(x) = 6x

. This changes sign on the interval

[-1, 1]

, so the estimate is not guaranteed to be an overestimate or underestimate.

Question 6

Let $f(x)$ be an even function such that $\int_0^2 f(x) dx$ is approximated by the trapezoidal rule with $n=2$ as $T_2 = 10$ . What is the trapezoidal rule approximation for $\int_{-2}^2 f(x) dx$ with $n=4$ ?

10
20 + f(0)
20 - f(0)
20 (correct answer)

Explanation: When you encounter trapezoidal rule questions involving even functions, remember that even functions have the special property

f(-x) = f(x)

, which creates useful symmetries you can exploit. Given that

f(x)

is even and

T_2 = 10

for

\int_0^2 f(x) dx

with

n=2

, let's find the trapezoidal approximation for

\int_{-2}^2 f(x) dx

with

n=4

. For the trapezoidal rule with

n=2

[0,2]

, we have

\Delta x = 1

and evaluation points at

x = 0, 1, 2

. So

T_2 = \frac{1}{2}[f(0) + 2f(1) + f(2)] = 10

. For

\int_{-2}^2 f(x) dx

with

n=4

, we have

\Delta x = 1

and evaluation points at

x = -2, -1, 0, 1, 2

. The trapezoidal rule gives:

T_4 = \frac{1}{2}[f(-2) + 2f(-1) + 2f(0) + 2f(1) + f(2)]

. Since

f(x)

is even,

f(-2) = f(2)

and

f(-1) = f(1)

. Substituting:

T_4 = \frac{1}{2}[f(2) + 2f(1) + 2f(0) + 2f(1) + f(2)] = \frac{1}{2}[2f(2) + 4f(1) + 2f(0)] = 2 \cdot \frac{1}{2}[f(0) + 2f(1) + f(2)] = 2T_2 = 20

. Choice A (10) incorrectly assumes the integral doesn't change when extending the domain. Choice B (20 + f(0)) adds an extra

f(0)

term. Choice C (20 - f(0)) subtracts

f(0)

incorrectly. Choice D (20) correctly recognizes that the integral doubles due to symmetry. Strategy tip: For even functions on symmetric intervals, the trapezoidal approximation on

[-a,a]

equals twice the approximation on

[0,a]

when using appropriate subdivisions.

Question 7

$R_n < T_n < I < L_n$ (correct answer)
$L_n < I < T_n < R_n$
$T_n < R_n < I < L_n$
$R_n < I < L_n < T_n$

L_n

) will be an overestimate, and the right-hand Riemann sum (

R_n

) will be an underestimate. This gives us the inequality

R_n < I < L_n

T_n

) provides an underestimate, so

T_n < I

.\n3. Comparing $R_n$ and $T_n$ : The formula for the trapezoidal rule is the average of the left and right Riemann sums:

T_n = \frac{L_n + R_n}{2}

. Since

R_n < L_n

, it follows that

R_n < \frac{L_n + R_n}{2} < L_n

, which means

R_n < T_n < L_n

.\nCombining these facts: We know

R_n < T_n

and

T_n < I

and

I < L_n

. Putting them all together gives the complete ordering:

R_n < T_n < I < L_n

Question 8

32 (correct answer)
35
36
38

Explanation: We use the linearity property of the trapezoidal rule approximation. Let

T_n(f)

denote the approximation for a function

f

.\n

T_n(ag+bh) = aT_n(g) + bT_n(h)

for constants

a, b

.\nWe want to find

T_4

for

\int_0^4 g(x) dx = \int_0^4 (3f(x) - 1) dx

.\nUsing the linearity property:\n

T_4(g) = T_4(3f - 1) = 3T_4(f) - T_4(1)

.\nWe are given

T_4(f) = 12

.\nWe need to calculate

T_4(1)

, which is the trapezoidal approximation for

\int_0^4 1 dx

. Since the function

h(x)=1

is linear, the trapezoidal rule is exact.\n

\int_0^4 1 dx = [x]_0^4 = 4

.\nSo,

T_4(1) = 4

.\nSubstituting these values back:\n

T_4(g) = 3(12) - 4 = 36 - 4 = 32

.\nA) This is the correct value.\nB) This results from

3(12)-1=35

, incorrectly subtracting the constant 1 instead of its integral approximation.\nC) This results from

3(12)=36

, ignoring the constant term -1 entirely.\nD) This results from

3(12)+2=38

, possibly making a sign error and a calculation error on the constant part.

Question 9

The approximation for $\int_1^3 f(x) dx$ using the trapezoidal rule with $n=2$ is 12. The approximation for the same integral with $n=4$ is 10. Given that the approximation $T_{2n}$ can be written as $T_{2n} = \frac{1}{2} T_n + M$ , where $M$ is a sum of midpoint function values, what is the value of $f(1.5)+f(2.5)$ ?

4
8 (correct answer)
10
16

Explanation: For an integral on

[a,b]

, the relationship between

T_n

and

T_{2n}

is given by

T_{2n} = \frac{1}{2}T_n + h_{2n} \sum_{k=1}^{n} f(x_{2k-1})

, where

h_{2n}

is the new step size and the sum is over the new midpoints.\nIn this problem, we are considering

\int_1^3 f(x) dx

T_2 = 12

and

T_4 = 10

. Here

n=2

, so

2n=4

.\nThe relationship is

T_4 = \frac{1}{2}T_2 + h_4 (f(x_1) + f(x_3))

.\nThe interval for

n=4

[1,3]

, so

h_4 = (3-1)/4 = 0.5

. The points are

x_0=1, x_1=1.5, x_2=2, x_3=2.5, x_4=3

. The new midpoints introduced when going from

n=2

n=4

are

x_1=1.5

and

x_3=2.5

.\nPlugging in the given values:\n

10 = \frac{1}{2}(12) + 0.5(f(1.5) + f(2.5))

10 = 6 + 0.5(f(1.5) + f(2.5))

4 = 0.5(f(1.5) + f(2.5))

8 = f(1.5) + f(2.5)

Question 10

Let $T_n(f)$ denote the trapezoidal rule approximation for $\int_a^b f(x) dx$ with $n$ subintervals. Given that $T_4$ for $\int_1^3 x^2 dx$ is $\frac{35}{4}$ , what is the value of $T_4$ for $\int_1^3 (x^2 - 2) dx$ ?

$\frac{3}{4}$
$\frac{19}{4}$ (correct answer)
$\frac{27}{4}$
$\frac{43}{4}$

Explanation: The trapezoidal rule is a linear operator, meaning

T_n(f+g) = T_n(f) + T_n(g)

and

T_n(cf) = cT_n(f)

for a constant

c

.\nTherefore,

T_4(\int_1^3 (x^2 - 2) dx) = T_4(\int_1^3 x^2 dx) - T_4(\int_1^3 2 dx)

.\nWe are given

T_4(\int_1^3 x^2 dx) = \frac{35}{4}

.\nThe trapezoidal rule is exact for linear functions, including constant functions. So, the approximation for

\int_1^3 2 dx

is equal to its exact value.\nExact value of

\int_1^3 2 dx = [2x]_1^3 = 2(3) - 2(1) = 6 - 2 = 4

.\nSo,

T_4(\int_1^3 2 dx) = 4

.\nThe required value is

\frac{35}{4} - 4 = \frac{35 - 16}{4} = \frac{19}{4}

.\nA) This results from subtracting

2 \times n = 2 \times 4 = 8

instead of 4.\nC) This results from simply subtracting 2 from the given value:

\frac{35}{4} - 2 = \frac{27}{4}

. This incorrectly assumes the integral of the constant is the constant value itself.\nD) This results from incorrectly adding 2 instead of subtracting 4.

Question 11

The length of the curve $y = \ln(\cos(x))$ from $x=0$ to $x=\pi/3$ is given by $L = \int_0^{\pi/3} \sec(x) dx$ . Find the trapezoidal rule approximation of $L$ with $n=2$ .

$\frac{\pi(9+2\sqrt{3})}{36}$
$\frac{\pi(3+2\sqrt{3})}{24}$
$\frac{\pi(9+4\sqrt{3})}{36}$ (correct answer)
$\frac{\pi(9+4\sqrt{3})}{18}$

Explanation: We need to approximate

\int_0^{\pi/3} \sec(x) dx

with

n=2

. The interval width is

h = \frac{\pi/3 - 0}{2} = \frac{\pi}{6}

. The x-values are

0, \pi/6, \pi/3

.\nLet

f(x) = \sec(x)

. We evaluate the function at these points:\n

f(0) = \sec(0) = 1

f(\pi/6) = \sec(\pi/6) = \frac{1}{\cos(\pi/6)} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}

f(\pi/3) = \sec(\pi/3) = \frac{1}{\cos(\pi/3)} = \frac{1}{1/2} = 2

\nNow apply the trapezoidal rule formula:\n

T_2 = \frac{h}{2}[f(0) + 2f(\pi/6) + f(\pi/3)]

T_2 = \frac{\pi/6}{2}[1 + 2(\frac{2\sqrt{3}}{3}) + 2] = \frac{\pi}{12}[3 + \frac{4\sqrt{3}}{3}]

T_2 = \frac{\pi}{12} \left( \frac{9+4\sqrt{3}}{3} \right) = \frac{\pi(9+4\sqrt{3})}{36}

.\nA) This results from forgetting to multiply the

2

from

2f(\pi/6)

with the numerator of

\frac{2\sqrt{3}}{3}

.\nB) This results from using

\cos(x)

values instead of

\sec(x)

for the middle and end points.\nD) This results from using

h

instead of

h/2

in the formula.

Question 12

The integral $I = \int_1^5 \frac{1}{x} dx$ is approximated using the trapezoidal rule with $n=2$ subintervals. What is the value of this approximation?

$\frac{14}{15}$
$\frac{56}{15}$
$\frac{32}{15}$
$\frac{28}{15}$ (correct answer)

Explanation: When you encounter integral approximation problems, you're applying numerical methods to estimate definite integrals. The trapezoidal rule is particularly common on IB exams because it's conceptually straightforward and gives reasonable approximations. The trapezoidal rule formula is

\frac{h}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n)]

, where

h = \frac{b-a}{n}

is the width of each subinterval. For this problem, you have

a = 1

b = 5

n = 2

, so

h = \frac{5-1}{2} = 2

. The x-values are

x_0 = 1

x_1 = 3

x_2 = 5

. Since

f(x) = \frac{1}{x}

, you get:

$f(1) = 1$
$f(3) = \frac{1}{3}$
$f(5) = \frac{1}{5}$

Applying the formula:

\frac{2}{2}[1 + 2(\frac{1}{3}) + \frac{1}{5}] = 1 + \frac{2}{3} + \frac{1}{5}

Converting to a common denominator of 15:

\frac{15}{15} + \frac{10}{15} + \frac{3}{15} = \frac{28}{15}

This confirms answer D is correct. Answer A (

\frac{14}{15}

) results from forgetting the factor of 2 in front of the middle term. Answer B (

\frac{56}{15}

) comes from incorrectly doubling the entire result. Answer C (

\frac{32}{15}

) appears when you mistakenly double both endpoint values instead of just the middle term. Remember: in the trapezoidal rule, only the interior points get multiplied by 2, never the endpoints. Always double-check your arithmetic when working with fractions.

Question 13

The trapezoidal rule is used to approximate $\int_2^6 f(x) dx$ with 4 subintervals. The values of the function at the relevant points are $f(2)=10, f(3)=k, f(4)=18, f(5)=22, f(6)=20$ . If the approximation is 70, what is the value of $k$ ?

12
14
15 (correct answer)
30

Explanation: The interval is

[2, 6]

with

n=4

, so the width of each subinterval is

h = \frac{6-2}{4} = 1

. The x-values are 2, 3, 4, 5, 6.\nThe trapezoidal rule formula is

T_4 = \frac{h}{2}[f(2) + 2f(3) + 2f(4) + 2f(5) + f(6)]

.\nWe are given

T_4 = 70

. Plugging in the values:\n

70 = \frac{1}{2}[10 + 2k + 2(18) + 2(22) + 20]

\nMultiply by 2:\n

140 = 10 + 2k + 36 + 44 + 20

140 = 110 + 2k

30 = 2k

k = 15

.\nA) and B) Result from arithmetic errors in the calculation.\nD) This results from forgetting to multiply

k

by 2 in the trapezoidal formula:

140 = 10 + k + 36 + 44 + 20 = 110 + k

, giving

k = 30

Question 14

4
8 (correct answer)
10
16

Explanation: For an integral on

[a,b]

, the relationship between

T_n

and

T_{2n}

is given by

T_{2n} = \frac{1}{2}T_n + h_{2n} \sum_{k=1}^{n} f(x_{2k-1})

, where

h_{2n}

is the new step size and the sum is over the new midpoints.\nIn this problem, we are considering

\int_1^3 f(x) dx

T_2 = 12

and

T_4 = 10

. Here

n=2

, so

2n=4

.\nThe relationship is

T_4 = \frac{1}{2}T_2 + h_4 (f(x_1) + f(x_3))

.\nThe interval for

n=4

[1,3]

, so

h_4 = (3-1)/4 = 0.5

. The points are

x_0=1, x_1=1.5, x_2=2, x_3=2.5, x_4=3

. The new midpoints introduced when going from

n=2

n=4

are

x_1=1.5

and

x_3=2.5

.\nPlugging in the given values:\n

10 = \frac{1}{2}(12) + 0.5(f(1.5) + f(2.5))

10 = 6 + 0.5(f(1.5) + f(2.5))

4 = 0.5(f(1.5) + f(2.5))

8 = f(1.5) + f(2.5)

Question 15

12
14
15 (correct answer)
30

Explanation: The interval is

[2, 6]

with

n=4

, so the width of each subinterval is

h = \frac{6-2}{4} = 1

. The x-values are 2, 3, 4, 5, 6.\nThe trapezoidal rule formula is

T_4 = \frac{h}{2}[f(2) + 2f(3) + 2f(4) + 2f(5) + f(6)]

.\nWe are given

T_4 = 70

. Plugging in the values:\n

70 = \frac{1}{2}[10 + 2k + 2(18) + 2(22) + 20]

\nMultiply by 2:\n

140 = 10 + 2k + 36 + 44 + 20

140 = 110 + 2k

30 = 2k

k = 15

.\nA) and B) Result from arithmetic errors in the calculation.\nD) This results from forgetting to multiply

k

by 2 in the trapezoidal formula:

140 = 10 + k + 36 + 44 + 20 = 110 + k

, giving

k = 30

Question 16

A table of values for a function $g(x)$ is given below.

x	2	3	4	5	6
g(x)	4	7	11	16	22

Use the trapezoidal rule with 4 subintervals to estimate $\int_0^4 g(x+2) dx$ .

30
47 (correct answer)
52
94

Explanation: First, we perform a substitution to simplify the integral. Let

u = x+2

. Then

du = dx

. We must also change the limits of integration: when

x=0

u=2

; when

x=4

u=6

. The integral becomes

\int_2^6 g(u) du

.\nNow we use the trapezoidal rule to estimate this new integral with

n=4

subintervals, using the values from the table. The interval is

[2, 6]

, so the width of each subinterval is

h = \frac{6-2}{4} = 1

. The u-values are 2, 3, 4, 5, 6, which match the table.\n

T_4 = \frac{h}{2}[g(2) + 2g(3) + 2g(4) + 2g(5) + g(6)]

T_4 = \frac{1}{2}[4 + 2(7) + 2(11) + 2(16) + 22]

T_4 = \frac{1}{2}[4 + 14 + 22 + 32 + 22]

T_4 = \frac{1}{2}[94] = 47

.\nA) This results from forgetting to multiply the middle terms by 2:

\frac{1}{2}(4+7+11+16+22) = 30

.\nC) This is the approximation with only one interval (n=1):

\frac{6-2}{2}(g(2)+g(6)) = 2(4+22) = 52

.\nD) This is the sum inside the brackets, forgetting the factor of

\frac{1}{2}

Question 17

$T_2 = \frac{1}{2}T_1 + 2f(1)$
$T_2 = \frac{1}{2}(T_1 + f(1))$
$T_2 = T_1 + f(1)$
$T_2 = \frac{1}{2}T_1 + f(1)$ (correct answer)

Explanation: First, let's write out the expressions for

T_1

and

T_2

.\nFor

T_1

, we have

n=1

[0, 2]

, so

h=2

. The points are

x_0=0, x_1=2

.\n

T_1 = \frac{h}{2}[f(0) + f(2)] = \frac{2}{2}[f(0) + f(2)] = f(0) + f(2)

.\nFor

T_2

, we have

n=2

[0, 2]

, so

h=1

. The points are

x_0=0, x_1=1, x_2=2

.\n

T_2 = \frac{h}{2}[f(0) + 2f(1) + f(2)] = \frac{1}{2}[f(0) + 2f(1) + f(2)]

.\nNow we substitute the expression for

T_1

into the expression for

T_2

. We can rewrite the

T_2

expression as:\n

T_2 = \frac{1}{2}[(f(0) + f(2)) + 2f(1)]

.\nSince

T_1 = f(0) + f(2)

, we can substitute it in:\n

T_2 = \frac{1}{2}[T_1 + 2f(1)] = \frac{1}{2}T_1 + f(1)

.\nDistractors A, B, and C are incorrect algebraic manipulations of the relationship.

Question 18

10
20 + f(0)
20 - f(0)
20 (correct answer)

Explanation: When you encounter trapezoidal rule questions involving even functions, remember that even functions have the special property

f(-x) = f(x)

, which creates useful symmetries you can exploit. Given that

f(x)

is even and

T_2 = 10

for

\int_0^2 f(x) dx

with

n=2

, let's find the trapezoidal approximation for

\int_{-2}^2 f(x) dx

with

n=4

. For the trapezoidal rule with

n=2

[0,2]

, we have

\Delta x = 1

and evaluation points at

x = 0, 1, 2

. So

T_2 = \frac{1}{2}[f(0) + 2f(1) + f(2)] = 10

. For

\int_{-2}^2 f(x) dx

with

n=4

, we have

\Delta x = 1

and evaluation points at

x = -2, -1, 0, 1, 2

. The trapezoidal rule gives:

T_4 = \frac{1}{2}[f(-2) + 2f(-1) + 2f(0) + 2f(1) + f(2)]

. Since

f(x)

is even,

f(-2) = f(2)

and

f(-1) = f(1)

. Substituting:

T_4 = \frac{1}{2}[f(2) + 2f(1) + 2f(0) + 2f(1) + f(2)] = \frac{1}{2}[2f(2) + 4f(1) + 2f(0)] = 2 \cdot \frac{1}{2}[f(0) + 2f(1) + f(2)] = 2T_2 = 20

. Choice A (10) incorrectly assumes the integral doesn't change when extending the domain. Choice B (20 + f(0)) adds an extra

f(0)

term. Choice C (20 - f(0)) subtracts

f(0)

incorrectly. Choice D (20) correctly recognizes that the integral doubles due to symmetry. Strategy tip: For even functions on symmetric intervals, the trapezoidal approximation on

[-a,a]

equals twice the approximation on

[0,a]

when using appropriate subdivisions.

Question 19

$\frac{3}{4}$
$\frac{19}{4}$ (correct answer)
$\frac{27}{4}$
$\frac{43}{4}$

Explanation: The trapezoidal rule is a linear operator, meaning

T_n(f+g) = T_n(f) + T_n(g)

and

T_n(cf) = cT_n(f)

for a constant

c

.\nTherefore,

T_4(\int_1^3 (x^2 - 2) dx) = T_4(\int_1^3 x^2 dx) - T_4(\int_1^3 2 dx)

.\nWe are given

T_4(\int_1^3 x^2 dx) = \frac{35}{4}

.\nThe trapezoidal rule is exact for linear functions, including constant functions. So, the approximation for

\int_1^3 2 dx

is equal to its exact value.\nExact value of

\int_1^3 2 dx = [2x]_1^3 = 2(3) - 2(1) = 6 - 2 = 4

.\nSo,

T_4(\int_1^3 2 dx) = 4

.\nThe required value is

\frac{35}{4} - 4 = \frac{35 - 16}{4} = \frac{19}{4}

.\nA) This results from subtracting

2 \times n = 2 \times 4 = 8

instead of 4.\nC) This results from simply subtracting 2 from the given value:

\frac{35}{4} - 2 = \frac{27}{4}

. This incorrectly assumes the integral of the constant is the constant value itself.\nD) This results from incorrectly adding 2 instead of subtracting 4.

Question 20

The approximation of $\int_1^5 f(x) dx$ using the trapezoidal rule with 4 subintervals is 34. Given the values $f(1)=3, f(2)=5, f(4)=11, f(5)=13$ , find the value of $f(3)$ .

-7
4
10 (correct answer)
20

Explanation: The trapezoidal rule formula is

T_n = \frac{h}{2}[f(x_0) + 2f(x_1) + ... + 2f(x_{n-1}) + f(x_n)]

.\nFor this problem,

a=1, b=5, n=4

, so the width of each subinterval is

h = \frac{5-1}{4} = 1

. The x-values are

x_0=1, x_1=2, x_2=3, x_3=4, x_4=5

.\nWe are given

T_4 = 34

. Plugging the values into the formula:\n

34 = \frac{1}{2}[f(1) + 2f(2) + 2f(3) + 2f(4) + f(5)]

68 = [3 + 2(5) + 2f(3) + 2(11) + 13]

68 = 3 + 10 + 2f(3) + 22 + 13

68 = 48 + 2f(3)

20 = 2f(3)

f(3) = 10

.\nA) This results from forgetting the

\frac{1}{2}

factor:

34 = 48 + 2f(3) \Rightarrow -14 = 2f(3) \Rightarrow f(3)=-7

.\nB) This results from misapplying the formula, such as not multiplying

f(2)

f(5)

by 2.\nD) This results from forgetting to multiply

f(3)

by 2:

68 = 48 + f(3) \Rightarrow f(3)=20