Riemann Integration Questions and Answers

Q: Total variation of the function :\[ f(x) = \frac{x^{3}}{3} - x, \ x \in [2, 3] \]is :

The total variation is 16/3.For a differentiable function, the total variation on \([a, b]\) is \(\int_{a}^{b} |f'(x)|\, dx\). Here \(f'(x) = x^{2} - 1\), which is positive on [2, 3], so f is increasing and the variation is simply \(f(3) - f(2)\):\[ \begin{aligned} f(3) &= 9 - 3 = 6 \\ f(2) &= \tfrac{8}{3} - 2 = \tfrac{2}{3} \\ V &= 6 - \tfrac{2}{3} = \tfrac{16}{3} \end{aligned} \]0 would hold only for a constant function. 17/3 and 13/3 come from slips in \(f(2)\), for example taking it as \(1/3\) or \(5/3\).

Q: The improper integral \(\displaystyle\int_{a}^{b} \frac{dx}{(x - a)^{n}}\) converges :

The keyed answer is only if \(0 \lt n \lt 1\).The integrand \(1/(x - a)^{n}\) blows up at the lower limit only when \(n \gt 0\), so that is when the integral is improper. For \(n \ne 1\):\[ \begin{aligned} &\int_{a+\epsilon}^{b} \frac{dx}{(x - a)^{n}} \\ &= \frac{(b - a)^{1-n} - \epsilon^{1-n}}{1 - n} \end{aligned} \]As \(\epsilon \to 0\), \(\epsilon^{1-n} \to 0\) if \(n \lt 1\) and \(\to \infty\) if \(n \gt 1\); for \(n = 1\) the integral is \(\log\) and diverges. So the improper integral converges exactly for \(0 \lt n \lt 1\). For \(n \le 0\) the integrand is continuous and the integral is an ordinary one, which is why the key states the range for the improper case.It does not converge for every \(n\), it diverges for \(n \gt 1\), and integers \(n \ge 1\) all diverge.

Q: Consider the following statements :I. A function \(f\) on \([a, b]\) is of bounded variation if, and only if, \(f\) is the difference of finite-valued monotone increasing functions on \([a, b]\).II. If the function \(f\) is of bounded variation on \([a, b]\), then \(f\) is measurable.

Both I and II are true.I is the Jordan decomposition theorem: a function of bounded variation on \([a, b]\) can be written as \(f = g - h\), with \(g(x) = V_{a}^{x}f\) (the total variation up to \(x\)) and \(h = g - f\), both monotone increasing and finite. Conversely, a difference of two finite monotone functions has bounded variation, since each has variation equal to its total rise.II follows from I: a monotone function is Borel measurable (the set where it exceeds a value is an interval), and the difference of two measurable functions is measurable. So every option saying one of them is false is wrong.

Q: The average of function f(x) = sin x in the interval 0 to π is :

The average is \(2/\pi\).The mean value of a function over an interval is its integral divided by the length of the interval:\[ \begin{aligned} \bar{f} &= \frac{1}{\pi - 0}\int_{0}^{\pi} \sin x\, dx \\ &= \frac{1}{\pi}\big[-\cos x\big]_{0}^{\pi} \\ &= \frac{1}{\pi}(1 + 1) = \frac{2}{\pi} \end{aligned} \]That is about 0.64. The value \(1/\pi\) forgets that \(-\cos x\) changes by 2, not 1, and \(4/\pi\) doubles the result. The value 1/2 is the average of \(\sin^{2} x\), the quantity behind RMS values, not of \(\sin x\).

Q: Let \(f : [0, 1] \to \mathbf{R}\) be continuous and \(\int_{0}^{x} f = \int_{x}^{1} f\) for all \(x \in [0, 1]\). Then :

\(f\) is constant (in fact identically zero).Write \(F(x) = \int_{0}^{x} f\) and \(I = \int_{0}^{1} f\). The condition says \(F(x) = I - F(x)\), so \(F(x) = I/2\) for every \(x\). Putting \(x = 0\) gives \(I/2 = F(0) = 0\), so \(F \equiv 0\).Since \(f\) is continuous, \(F' = f\) by the fundamental theorem of calculus, so \(f \equiv 0\), a constant function.\(f(t) = t\) fails: \(\int_{0}^{x} t\,dt = x^{2}/2\) but \(\int_{x}^{1} t\,dt = (1 - x^{2})/2\). \(t^{2}\) fails likewise, and a strictly monotonic function cannot be identically zero.

Q: Suppose \(f(x)\) is defined on [0, 1] as follows :\[ f(x) = \begin{cases} 0, & \text{if } x \text{ irrational} \\ 1/q, & \text{if } x \text{ rational} \end{cases} \]and \(x = \dfrac{p}{q}\) with \((p, q) = 1\). Then \(f\) is :

Thomae's function is Riemann integrable (its integral is 0).For any \(\varepsilon \gt 0\), only finitely many rationals \(p/q\) in [0, 1] have \(1/q \ge \varepsilon\). Cover them by intervals of tiny total length; on the rest of [0, 1] the function is below \(\varepsilon\). So upper sums can be made less than \(2\varepsilon\), while every lower sum is 0 because irrationals lie in every subinterval.The function is continuous at every irrational point (values near an irrational have large denominators, so they are small) and discontinuous at every rational point (where \(f = 1/q \gt 0\) but irrationals arbitrarily close give 0). That rules out (B), (C) and (D).

Q: Which of the following statements holds for a function \(f : [0, 1] \to \mathbf{R}\) ?

The statement that holds is that if \(f\) is of bounded variation, then \(f\) is Riemann integrable.A function of bounded variation on \([0, 1]\) is the difference of two increasing functions. Increasing functions are bounded and have at most countably many discontinuities, a set of measure zero, so by Lebesgue's criterion they are Riemann integrable, and so is their difference.The other implications fail. A step function is of bounded variation but not continuous, let alone uniformly continuous (B). The Dirichlet function, 1 on rationals and 0 elsewhere, is bounded but not of bounded variation (C). And \(x\sin(1/x)\) with value 0 at 0 is uniformly continuous on \([0, 1]\) but of unbounded variation (D).

Q: Suppose a bounded function \(f : [0, 1] \to \mathbf{R}\) is Riemann integrable. Then which of the following is not true ?

The statement that is not true is f is not Lebesgue integrable whenever f is discontinuous.Every bounded Riemann integrable function on [0, 1] is Lebesgue integrable, and its two integrals agree, so the first two statements are true. Since \(f\) is Lebesgue integrable, so is \(|f|\), which makes the last statement true as well.Discontinuity does not stop Lebesgue integrability: a step function, or any Riemann integrable function with a discontinuity, is Lebesgue integrable. Even the Dirichlet function, discontinuous everywhere, is Lebesgue integrable with integral 0.

11.

Total variation of the function :

\[ f(x) = \frac{x^{3}}{3} - x, \ x \in [2, 3] \]

is :

A.16/3
B.17/3
C.0
D.13/3

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The total variation is 16/3.

For a differentiable function, the total variation on \([a, b]\) is \(\int_{a}^{b} |f'(x)|\, dx\). Here \(f'(x) = x^{2} - 1\), which is positive on [2, 3], so f is increasing and the variation is simply \(f(3) - f(2)\):

\[ \begin{aligned} f(3) &= 9 - 3 = 6 \\ f(2) &= \tfrac{8}{3} - 2 = \tfrac{2}{3} \\ V &= 6 - \tfrac{2}{3} = \tfrac{16}{3} \end{aligned} \]

0 would hold only for a constant function. 17/3 and 13/3 come from slips in \(f(2)\), for example taking it as \(1/3\) or \(5/3\).

Discuss in Forum

12.

The improper integral \(\displaystyle\int_{a}^{b} \frac{dx}{(x - a)^{n}}\) converges :

A.\(\forall\, n\)
B.if \(n \gt 1\)
C.if \(n\) is an integer
D.only if \(0 \lt n \lt 1\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The keyed answer is only if \(0 \lt n \lt 1\).

The integrand \(1/(x - a)^{n}\) blows up at the lower limit only when \(n \gt 0\), so that is when the integral is improper. For \(n \ne 1\):

\[ \begin{aligned} &\int_{a+\epsilon}^{b} \frac{dx}{(x - a)^{n}} \\ &= \frac{(b - a)^{1-n} - \epsilon^{1-n}}{1 - n} \end{aligned} \]

As \(\epsilon \to 0\), \(\epsilon^{1-n} \to 0\) if \(n \lt 1\) and \(\to \infty\) if \(n \gt 1\); for \(n = 1\) the integral is \(\log\) and diverges. So the improper integral converges exactly for \(0 \lt n \lt 1\). For \(n \le 0\) the integrand is continuous and the integral is an ordinary one, which is why the key states the range for the improper case.

It does not converge for every \(n\), it diverges for \(n \gt 1\), and integers \(n \ge 1\) all diverge.

Discuss in Forum

13.

Consider the following statements :

I. A function \(f\) on \([a, b]\) is of bounded variation if, and only if, \(f\) is the difference of finite-valued monotone increasing functions on \([a, b]\).
II. If the function \(f\) is of bounded variation on \([a, b]\), then \(f\) is measurable.

A.Only I is true
B.Only II is true
C.Both I and II are not true
D.Both I and II are true

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

Both I and II are true.

I is the Jordan decomposition theorem: a function of bounded variation on \([a, b]\) can be written as \(f = g - h\), with \(g(x) = V_{a}^{x}f\) (the total variation up to \(x\)) and \(h = g - f\), both monotone increasing and finite. Conversely, a difference of two finite monotone functions has bounded variation, since each has variation equal to its total rise.

II follows from I: a monotone function is Borel measurable (the set where it exceeds a value is an interval), and the difference of two measurable functions is measurable. So every option saying one of them is false is wrong.

Discuss in Forum

14.

The average of function f(x) = sin x in the interval 0 to π is :

A.1/2
B.2/π
C.1/π
D.4/π

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The average is \(2/\pi\).

The mean value of a function over an interval is its integral divided by the length of the interval:

\[ \begin{aligned} \bar{f} &= \frac{1}{\pi - 0}\int_{0}^{\pi} \sin x\, dx \\ &= \frac{1}{\pi}\big[-\cos x\big]_{0}^{\pi} \\ &= \frac{1}{\pi}(1 + 1) = \frac{2}{\pi} \end{aligned} \]

That is about 0.64. The value \(1/\pi\) forgets that \(-\cos x\) changes by 2, not 1, and \(4/\pi\) doubles the result. The value 1/2 is the average of \(\sin^{2} x\), the quantity behind RMS values, not of \(\sin x\).

Discuss in Forum

15.

Let \(f : [0, 1] \to \mathbf{R}\) be continuous and \(\int_{0}^{x} f = \int_{x}^{1} f\) for all \(x \in [0, 1]\). Then :

A.\(f\) is strictly monotonic
B.\(f\) is constant
C.\(f(t) = t\) for all \(t \in [0, 1]\)
D.\(f(t) = t^{2}\) for all \(t \in [0, 1]\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

\(f\) is constant (in fact identically zero).

Write \(F(x) = \int_{0}^{x} f\) and \(I = \int_{0}^{1} f\). The condition says \(F(x) = I - F(x)\), so \(F(x) = I/2\) for every \(x\). Putting \(x = 0\) gives \(I/2 = F(0) = 0\), so \(F \equiv 0\).

Since \(f\) is continuous, \(F' = f\) by the fundamental theorem of calculus, so \(f \equiv 0\), a constant function.

\(f(t) = t\) fails: \(\int_{0}^{x} t\,dt = x^{2}/2\) but \(\int_{x}^{1} t\,dt = (1 - x^{2})/2\). \(t^{2}\) fails likewise, and a strictly monotonic function cannot be identically zero.

Discuss in Forum

16.

Suppose \(f(x)\) is defined on [0, 1] as follows :

\[ f(x) = \begin{cases} 0, & \text{if } x \text{ irrational} \\ 1/q, & \text{if } x \text{ rational} \end{cases} \]

and \(x = \dfrac{p}{q}\) with \((p, q) = 1\). Then \(f\) is :

A.Riemann integrable
B.Continuous at rational points
C.Discontinuous at irrational points
D.Discontinuous everywhere

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

Thomae's function is Riemann integrable (its integral is 0).

For any \(\varepsilon \gt 0\), only finitely many rationals \(p/q\) in [0, 1] have \(1/q \ge \varepsilon\). Cover them by intervals of tiny total length; on the rest of [0, 1] the function is below \(\varepsilon\). So upper sums can be made less than \(2\varepsilon\), while every lower sum is 0 because irrationals lie in every subinterval.

The function is continuous at every irrational point (values near an irrational have large denominators, so they are small) and discontinuous at every rational point (where \(f = 1/q \gt 0\) but irrationals arbitrarily close give 0). That rules out (B), (C) and (D).

Discuss in Forum

17.

Which of the following statements holds for a function \(f : [0, 1] \to \mathbf{R}\) ?

A.If \(f\) is of bounded variation, then \(f\) is Riemann integrable
B.If \(f\) is of bounded variation, then it is uniformly continuous
C.If \(f\) is bounded, then it is of bounded variation
D.If \(f\) is uniformly continuous, then it is of bounded variation

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The statement that holds is that if \(f\) is of bounded variation, then \(f\) is Riemann integrable.

A function of bounded variation on \([0, 1]\) is the difference of two increasing functions. Increasing functions are bounded and have at most countably many discontinuities, a set of measure zero, so by Lebesgue's criterion they are Riemann integrable, and so is their difference.

The other implications fail. A step function is of bounded variation but not continuous, let alone uniformly continuous (B). The Dirichlet function, 1 on rationals and 0 elsewhere, is bounded but not of bounded variation (C). And \(x\sin(1/x)\) with value 0 at 0 is uniformly continuous on \([0, 1]\) but of unbounded variation (D).

Discuss in Forum

18.

Suppose a bounded function \(f : [0, 1] \to \mathbf{R}\) is Riemann integrable. Then which of the following is not true ?

A.\(f\) is Lebesgue integrable
B.Riemann integral of \(f\) is same as Lebesgue integral of \(f\)
C.\(f\) is not Lebesgue integrable whenever \(f\) is discontinuous
D.\(|f|\) is Lebesgue integrable

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The statement that is not true is f is not Lebesgue integrable whenever f is discontinuous.

Every bounded Riemann integrable function on [0, 1] is Lebesgue integrable, and its two integrals agree, so the first two statements are true. Since \(f\) is Lebesgue integrable, so is \(|f|\), which makes the last statement true as well.

Discontinuity does not stop Lebesgue integrability: a step function, or any Riemann integrable function with a discontinuity, is Lebesgue integrable. Even the Dirichlet function, discontinuous everywhere, is Lebesgue integrable with integral 0.

Riemann Integration Questions and Answers

Topics

Other Subjects

Tags: Riemann Integration Questions and Answers

About Us

Riemann Integration Questions and Answers

Topics

Other Subjects

Tags: Riemann Integration Questions and Answers

About Us

Stay informed about all important Dates & Notifications