The curves $y = x^{2}$ and $y = 4 x - 3$ intersect at two points. What is the area of the region bounded by these curves?

medium MCQ 3 marks Class 12 · Mathematics · Integrals

Answer

Option B:

\frac{4}{3}

square units

The correct answer is Option B: $\frac{4}{3}$ square units — Select the matching option.

Step-by-step solution

Set the equations equal to find intersection points: $x^{2} = 4 x - 3 \Rightarrow x^{2} - 4 x + 3 = 0$ .
Factor: $(x - 1) (x - 3) = 0$ , so $x = 1$ and $x = 3$ are the points of intersection.
For $1 \leq x \leq 3$ , the line $y = 4 x - 3$ lies above the parabola $y = x^{2}$ (check at $x = 2$ ).
Write the area integral: $A = \int_{1}^{3} [(4 x - 3) - x^{2}] d x$ .
Simplify the integrand: $- x^{2} + 4 x - 3$ .
Integrate: $\int (- x^{2} + 4 x - 3) d x = - \frac{x ^{3}}{3} + 2 x^{2} - 3 x$ .
Evaluate between the limits: at $x = 3$ , $- \frac{27}{3} + 18 - 9 = 0$ ; at $x = 1$ , $- \frac{1}{3} + 2 - 3 = - \frac{4}{3}$ .
Subtract lower limit from upper limit: $0 - (- \frac{4}{3}) = \frac{4}{3}$ .
Select the matching option.

Common mistake: Reversing the order of subtraction (upper‑minus‑lower) which gives a negative area.

Concept tested: Area between two curves