Q227 / 269 · NDA 2025 (Phase 1) · Mathematics
Consider the following for the two (02) items that follow:
The slope of the tangent to the curve $y = f(x)$ at $(x, f(x))$ is 4 for every real number $x$ and the curve passes through the origin.
What is the area bounded by the curve, the $x$-axis and the line $x = 4$?
The slope of the tangent to the curve $y = f(x)$ at $(x, f(x))$ is 4 for every real number $x$ and the curve passes through the origin.
What is the area bounded by the curve, the $x$-axis and the line $x = 4$?
- A$8$ square units
- B$16$ square units
- Correct answerC$32$ square units
- D$64$ square units
Solution
Given $\dfrac{dy}{dx} = 4$
Integrating both sides, $y = 4x + C$
Since the curve passes through the origin $(0,0)$, we have $0 = 4(0) + C \Rightarrow C = 0$
The equation of the curve is $y = 4x$
The area bounded by the curve, the $x$-axis and the line $x = 4$ is given by $\int_{0}^{4} y \, dx$
Area $= \int_{0}^{4} 4x \, dx = \left[ 2x^2 \right]_{0}^{4} = 2(16) - 0 = 32$ square units.
Answer: $32$ square units
Integrating both sides, $y = 4x + C$
Since the curve passes through the origin $(0,0)$, we have $0 = 4(0) + C \Rightarrow C = 0$
The equation of the curve is $y = 4x$
The area bounded by the curve, the $x$-axis and the line $x = 4$ is given by $\int_{0}^{4} y \, dx$
Area $= \int_{0}^{4} 4x \, dx = \left[ 2x^2 \right]_{0}^{4} = 2(16) - 0 = 32$ square units.
Answer: $32$ square units
Hence, the correct answer is option: 'C'.