Question

The area enclosed between the curve $y = 11x - 24 - x^2$ and the line $y = x$ is

• A. $\frac{1}{3}$
• B. $\frac{3}{4}$
• C. $1$
• D. $\frac{4}{3}$
• E. $\frac{1}{2}$

Hint:

Always subtract lower curve from upper curve in definite integrals.

Answer 1

The Correct Option is D

Step 1: Find intersection points \[ 11x - 24 - x^2 = x \] \[ -x^2 + 10x - 24 = 0 \] \[ x^2 - 10x + 24 = 0 \] \[ (x-4)(x-6)=0 \Rightarrow x=4,6 \]

Step 2: Area formula \[ \text{Area} = \int_{4}^{6} \left[(11x-24-x^2) - x\right] dx \] \[ = \int_{4}^{6} (10x - 24 - x^2) dx \]

Step 3: Integrate \[ = \left[5x^2 - 24x - \frac{x^3}{3}\right]_{4}^{6} \]

Step 4: Substitute limits At $x=6$: \[ 5(36) - 144 - 72 = 180 -144 -72 = -36 \] At $x=4$: \[ 5(16) - 96 - \frac{64}{3} = 80 -96 - \frac{64}{3} = -16 - \frac{64}{3} = -\frac{112}{3} \]

Step 5: Final area \[ = -36 + \frac{112}{3} = \frac{-108 + 112}{3} = \frac{4}{3} \] \[ \boxed{\frac{4}{3}} \]