Question

The area of the region \( \{(x,y): x^2-8x \le y \le -x\} \) is :

• A. \( \frac{343}{6} \)
• B. \( \frac{637}{6} \)
• C. \( \frac{437}{6} \)
• D. \( \frac{523}{6} \)

Answer 1

The Correct Option is A

Concept: The area between two curves \(y=f(x)\) and \(y=g(x)\) is \[ \text{Area}=\int_a^b [f(x)-g(x)]dx \] where \(f(x)\) is the upper curve and \(g(x)\) is the lower curve. Step 1: {Find the intersection points.} Given curves: \[ y=-x \] \[ y=x^2-8x \] Equating: \[ x^2-8x=-x \] \[ x^2-7x=0 \] \[ x(x-7)=0 \] \[ x=0,7 \] Step 2: {Identify upper and lower curves.} For \(x\in[0,7]\), \[ y=-x \] is above \[ y=x^2-8x \] Step 3: {Set up the integral.} \[ A=\int_0^7[(-x)-(x^2-8x)]dx \] \[ =\int_0^7(-x-x^2+8x)dx \] \[ =\int_0^7(7x-x^2)dx \] Step 4: {Evaluate the integral.} \[ \int_0^7(7x-x^2)dx \] \[ =\left[\frac{7x^2}{2}-\frac{x^3}{3}\right]_0^7 \] \[ =\frac{7\cdot49}{2}-\frac{343}{3} \] \[ =\frac{343}{2}-\frac{343}{3} \] \[ =343\left(\frac12-\frac13\right) \] \[ =\frac{343}{6} \]