Integration: Question 7

Worked solution

Part (a): Finding where the curve crosses the x-axis

Set $y=0$:

$x^2 - 6x + 5 = 0$

Factorise:

$(x-1)(x-5) = 0 \implies x = 1 \ \text{ or } \ x = 5$

Recompute independently by substituting back: at $x=1$, $y = 1-6+5=0$ ✓. At $x=5$, $y=25-30+5=0$ ✓. Both roots check out.

Part (b): Area enclosed between the curve and the x-axis

Since the curve lies below the $x$-axis on $[1,5]$, the raw definite integral of $y$ over this interval will come out negative, and the area is its magnitude.

$\int_1^5 (x^2-6x+5)\,dx$

Integrate:

$\int (x^2-6x+5)\,dx = \frac{x^3}{3} - 3x^2 + 5x$

Evaluate at the limits:

At $x=5$: $\ \dfrac{125}{3} - 3(25) + 5(5) = \dfrac{125}{3} - 75 + 25 = \dfrac{125}{3} - 50 = \dfrac{125-150}{3} = -\dfrac{25}{3}$

At $x=1$: $\ \dfrac{1}{3} - 3(1) + 5(1) = \dfrac{1}{3} - 3 + 5 = \dfrac{1}{3} + 2 = \dfrac{7}{3}$

$\int_1^5 (x^2-6x+5)\,dx = -\frac{25}{3} - \frac{7}{3} = -\frac{32}{3}$

The integral is negative because the curve is below the axis, so the area is the magnitude:

$\text{Area} = \left|-\frac{32}{3}\right| = \frac{32}{3}$

Recompute independently two ways:

1. Differentiate the antiderivative back: $\dfrac{d}{dx}\left(\dfrac{x^3}{3}-3x^2+5x\right) = x^2-6x+5$, which matches the integrand exactly, confirming the antiderivative is correct.
2. Use the “parabola hump” shortcut: for $-(x-p)(x-q)$ integrated between the roots $p$ and $q$, the enclosed area equals $\dfrac{(q-p)^3}{6}$. Here $x^2-6x+5=(x-1)(x-5)$ opens the same way relative to the axis, with $p=1$, $q=5$, so Area $=\dfrac{(5-1)^3}{6}=\dfrac{64}{6}=\dfrac{32}{3}$.

Both checks agree with the direct calculation, confirming the area is $\dfrac{32}{3}$.

Final answers

• (a) The curve crosses the $x$-axis at $x=1$ and $x=5$
• (b) Area enclosed $= \dfrac{32}{3}$ square units $\left(=10\dfrac{2}{3}\right)$