Let f be a continuous function on the intervals involved, where a and b are real numbers. Which of the following statements are correct?
(A) \(\int_{a}^{b}\displaystyle f(x)\,dx = \int_{a}^{b}\displaystyle f(a+b-x)\,dx\)
(B) \(\int_{0}^{2a}\displaystyle f(x)\,dx = 2\int_{0}^{a}\displaystyle f(x)\,dx\) for every continuous function f
(C) \(\int_{-a}^{a}\displaystyle f(x)\,dx = 0\) if f is an odd function
(D) \(\int_{a}^{b}\displaystyle f(x)\,dx = -\int_{b}^{a}\displaystyle f(x)\,dx\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (C) and (D) only
Replacing x by \(a+b-x\) maps the interval \([a, b]\) onto itself, so the integral does not change. Interchanging the limits changes the sign of the integral. For an odd function the parts on \([-a, 0]\) and \([0, a]\) cancel.
The rule \(\int_{0}^{2a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx\) holds only when \(f(2a-x) = f(x)\). For example, with \(f(x) = x\) and \(a = 1\):
\( \displaystyle \int_{0}^{2}x\,dx = 2, \qquad 2\int_{0}^{1}x\,dx = 1 \)
So the property \(\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a+b-x)\,dx\), the sign change on interchanging the limits and the zero integral of an odd function are always true, while the doubling rule needs \(f(2a-x) = f(x)\).
Checking the statements
(A) \(\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a+b-x)\,dx\) — Correct. The substitution \(t = a+b-x\) keeps the limits a and b.
(B) \(\int_{0}^{2a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx\) for every continuous function f — Incorrect. For \(f(x) = x\), \(a = 1\): the left side is 2, the right side is 1.
(C) \(\int_{-a}^{a} f(x)\,dx = 0\) if f is an odd function — Correct. The parts over \([-a, 0]\) and \([0, a]\) are equal and opposite.
(D) \(\int_{a}^{b} f(x)\,dx = -\int_{b}^{a} f(x)\,dx\) — Correct. Both sides equal \(F(b) - F(a)\), where F is an antiderivative of f.