Find the area bounded by the curve y = 6x² - 3x³

Find the area bounded by the curve $y = 6x^2 - 3x^3$, the $y$-axis, and the tangent to the curve drawn at the point $M(3, y)$.

We find the point of tangency, the tangent line, and then integrate the gap between the line and the curve.

Step 1: Find the point $M$

At $x=3$: $y = 6(3)^2-3(3)^3 = 54-81=-27$. So $M=(3,-27)$.

Step 2: Find the tangent line at $M$

\[\frac{dy}{dx}=12x-9x^2\]

At $x=3$: $m=12(3)-9(3)^2=36-81=-45$.

\[y-(-27) = -45(x-3) \implies y = -45x+108\]

Step 3: Find where the tangent and curve meet the $y$-axis

Tangent at $x=0$: $y=108$. Curve at $x=0$: $y=0$.

Step 4: Determine which curve is above on $[0,3]$

At $x=1$: curve gives $y=6-3=3$; tangent gives $y=-45+108=63$. Since $63>3$, the tangent lies above the cubic throughout $[0,3]$ (the tangent only touches the curve at $x=3$).

Step 5: Set up the integral

\[A = \int_0^3 \left[(-45x+108)-(6x^2-3x^3)\right]dx = \int_0^3 (3x^3-6x^2-45x+108)\,dx\]

Step 6: Evaluate the integral

\[A = \left[\frac{3x^4}{4}-2x^3-\frac{45x^2}{2}+108x\right]_0^3\]

At $x=3$: $\dfrac{3\cdot81}{4}-2\cdot27-\dfrac{45\cdot9}{2}+108\cdot3 = \dfrac{243}{4}-54-\dfrac{405}{2}+324$.

Converting to a common denominator of 4: $\dfrac{243-216-810+1296}{4}=\dfrac{513}{4}$.

At $x=0$: all terms vanish.

\[\boxed{A = \frac{513}{4} = 128.25 \text{ square units}}\]

Step 7: Verification of the arithmetic

\[243-216-810+1296 = 1539-1026 = 513 \quad\checkmark\]

Summary

\[\boxed{A = \frac{513}{4} = 128.25 \text{ square units}}\]

Key insight: Because the tangent point $x=3$ lies well past where the cubic $6x^2-3x^3$ has already turned sharply negative, the tangent line (with its steep slope of $-45$) separates from the curve dramatically over $[0,3]$, producing a large enclosed area despite the modest width of the interval.