Let R be the region bounded by the curve \(y = x^{3}\), the x-axis and the lines x = −2 and x = 2.
(A) \(\displaystyle\int_{-2}^{2} x^{3}\,dx = 0\)
(B) The area of R is 0 sq. units
(C) The area of R is 8 sq. units
(D) The area of R is equal to \(\displaystyle 2\int_{0}^{2} x^{3}\,dx\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C) and (D) only
The curve \(y = x^{3}\) is below the x-axis for x < 0 and above it for x > 0. Area is never negative, so the part below the axis is taken with its absolute value.
\( \displaystyle \int_{0}^{2} x^{3}\,dx = \left[\frac{x^{4}}{4}\right]_{0}^{2} = 4 \)
\( \displaystyle \int_{-2}^{0} x^{3}\,dx = \left[\frac{x^{4}}{4}\right]_{-2}^{0} = -4 \)
So the integral over the whole interval and the area are
\( \displaystyle \int_{-2}^{2} x^{3}\,dx = -4+4 = 0 \)
\(\text{Area of R} = |-4| + 4 = 8\)
So the integral from −2 to 2 is 0, the area of R is 8, and \(2\int_{0}^{2} x^{3}\,dx = 8\).
Checking the statements
(A) \(\displaystyle\int_{-2}^{2} x^{3}\,dx = 0\) — Correct. The two parts, −4 and 4, cancel.
(B) The area of R is 0 sq. units — Incorrect. The area is 4 + 4 = 8; the parts do not cancel in an area.
(C) The area of R is 8 sq. units — Correct. |−4| + 4 = 8.
(D) The area of R is equal to \(\displaystyle 2\int_{0}^{2} x^{3}\,dx\) — Correct. The two halves are equal in size, so the area is 2 × 4 = 8.