Let f be a continuous function on \([-3, 3]\) such that \(\int_{0}^{3}\displaystyle f(x)\,dx = 5\). Which of the following statements are correct?
(A) If f is an even function, then \(\int_{-3}^{3}\displaystyle f(x)\,dx = 10\)
(B) If f is an odd function, then \(\int_{-3}^{0}\displaystyle f(x)\,dx = -5\)
(C) \(\int_{0}^{3}\displaystyle f(3-x)\,dx = 5\)
(D) \(\int_{0}^{3}\displaystyle\left[f(x)+f(3-x)\right]dx = 5\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (B) and (C) only
Putting \(x = -t\) in the integral over \([-3, 0]\):
\( \displaystyle \int_{-3}^{0}f(x)\,dx = \int_{0}^{3}f(-t)\,dt \)
For an even function \(f(-t) = f(t)\), so this is 5 and \(\int_{-3}^{3} f = 5 + 5 = 10\).
For an odd function \(f(-t) = -f(t)\), so this is \(-5\).
By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx\),
\( \displaystyle \int_{0}^{3}f(3-x)\,dx = \int_{0}^{3}f(x)\,dx = 5 \)
\( \displaystyle \int_{0}^{3}\left[f(x)+f(3-x)\right]dx = 5 + 5 = 10 \)
So the four integrals are (A) 10, (B) −5, (C) 5 and (D) 10.
Checking the statements
(A) If f is an even function, then \(\int_{-3}^{3}f(x)\,dx = 10\) — Correct. \(2 \times 5 = 10\).
(B) If f is an odd function, then \(\int_{-3}^{0}f(x)\,dx = -5\) — Correct. \(\int_{0}^{3}f(-t)\,dt = -5\).
(C) \(\int_{0}^{3}f(3-x)\,dx = 5\) — Correct. Replacing x by \(3-x\) does not change the integral over [0, 3].
(D) \(\int_{0}^{3}\left[f(x)+f(3-x)\right]dx = 5\) — Incorrect. Each part is 5, so the sum is 10.