Question

The area (in sq. units) of the region bounded between the curve \( y=x^{2}+5x+1 \) and the line \( 7x-y+1=0 \) is

• A. \( 2 \)
• B. \( \frac{3}{4} \)
• C. \( \frac{4}{3} \)
• D. \( \frac{2}{5} \)

Hint:

Archimedes' Formula Shortcut: The area enclosed by a parabola and a line cutting it at two points is always equal to \( \frac{|a|}{6}(\beta - \alpha)^3 \). Here, \( \frac{1}{6}(2 - 0)^3 = \frac{8}{6} = \frac{4}{3} \). You can find the answer in under 5 seconds!

Answer 1

The Correct Option is C

Step 1: Find where the curve and line meet.
The line gives $y=7x+1$. Set equal to the curve $y=x^2+5x+1$: \[ 7x+1=x^2+5x+1\implies x^2-2x=0\implies x(x-2)=0 \] So $x=0$ and $x=2$.
Step 2: Decide which is on top.
Test $x=1$: line gives $y=8$, curve gives $y=7$. The line is higher, so it is the upper curve.
Step 3: Write the area integral.
\[ \text{Area}=\int_0^2[(7x+1)-(x^2+5x+1)]\,dx=\int_0^2(2x-x^2)\,dx \]
Step 4: Integrate.
\[ \int_0^2(2x-x^2)\,dx=\left[x^2-\frac{x^3}{3}\right]_0^2 \]
Step 5: Put in the limits.
\[ =\left(4-\frac{8}{3}\right)-0=\frac{12-8}{3} \]
Step 6: Simplify.
\[ =\frac{4}{3} \] \[ \boxed{\tfrac{4}{3}\text{ sq units}} \]