Question: Suppose that the area bounded by the curve $y = 3x^2 − 12x + 9$, the $x$-axis and the lines $x = 1$ and $x = k$ is $16$. Find $k$.
I tried but found no solution.
What I did:
$$\int_1^k (3x^2-12x+9) \,dx = \big[x^3-6x+9x \big]_1^k = k^3-6k^2+9k-4 =16$$ $$\Rightarrow k^3-6k^2+9k-20 = 0$$
Then after simplifying I got:
$$(k-5)(k^2-k+4) = 0$$
This means $k = 5$, because $(k^2-k+4)$ has no real roots, but this is wrong. The reason is after drawing the graph, the area under the curve, from $x=1$ to $x=5$ is:
$$-\int_1^3 (3x^2-12x+9) \, dx + \int_3^5 (3x^2-12x+9) \, dx = 24 \ne 16$$
Can anyone help? What mistake did I make?