Area Bounded by Polynomial Curve in Given Interval

To find the area bounded by the curve $y = f(x)$ and the lines $x = 0$, $y = 0$, and $x = t$, we first need to determine the value of $t = |s|$, where $s$ is the real root of $f(x) = 1 + 2x + 3x^2 + 4x^3$.

Step 1: Analyze the roots of $f(x)$

The given polynomial is:
$f(x) = 4x^3 + 3x^2 + 2x + 1$

Taking the first derivative with respect to $x$:
$f'(x) = 12x^2 + 6x + 2 = 2(6x^2 + 3x + 1)$

The discriminant of $6x^2 + 3x + 1$ is:
$\Delta = 3^2 - 4(6)(1) = 9 - 24 = -15 < 0$

Since $f'(x) > 0$ for all $x \in \mathbb{R}$, $f(x)$ is strictly increasing on $\mathbb{R}$. This implies that $f(x)$ has exactly one real root, $s$.

Step 2: Locate the real root $s$

We evaluate $f(x)$ at specific points:

• For $x = -\frac{1}{2}$:
  $f\left(-\frac{1}{2}\right) = 1 + 2\left(-\frac{1}{2}\right) + 3\left(-\frac{1}{2}\right)^2 + 4\left(-\frac{1}{2}\right)^3 = 1 - 1 + \frac{3}{4} - \frac{1}{2} = \frac{1}{4} > 0$

• For $x = -\frac{3}{4}$:
  $f\left(-\frac{3}{4}\right) = 1 + 2\left(-\frac{3}{4}\right) + 3\left(-\frac{3}{4}\right)^2 + 4\left(-\frac{3}{4}\right)^3 = 1 - \frac{3}{2} + \frac{27}{16} - \frac{27}{16} = -\frac{1}{2} < 0$

By the Intermediate Value Theorem, the unique real root $s$ lies in the interval:
$s \in \left(-\frac{3}{4}, -\frac{1}{2}\right)$

Therefore, $t = |s| = -s$ lies in the interval:
$t \in \left(\frac{1}{2}, \frac{3}{4}\right)$

Step 3: Calculate the bounded area

Since $t > 0$ and $f(x) > 0$ for all $x \ge 0$, the area $A$ bounded by $y = f(x)$, $x = 0$, $y = 0$, and $x = t$ is given by the integral:
$A = \int_{0}^{t} f(x) \, dx = \int_{0}^{t} (1 + 2x + 3x^2 + 4x^3) \, dx$

Evaluating the integral:
$A(t) = \left[ x + x^2 + x^3 + x^4 \right]_{0}^{t} = t + t^2 + t^3 + t^4$

Since $t > 0$, the function $A(t)$ is strictly increasing. We now evaluate $A(t)$ at the endpoints of the interval $\left(\frac{1}{2}, \frac{3}{4}\right)$:

• At $t = \frac{1}{2}$:
  $A\left(\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{8 + 4 + 2 + 1}{16} = \frac{15}{16}$

• At $t = \frac{3}{4}$:
  $A\left(\frac{3}{4}\right) = \frac{3}{4} + \frac{9}{16} + \frac{27}{64} + \frac{81}{256} = \frac{192 + 144 + 108 + 81}{256} = \frac{525}{256}$

Thus, the area $A$ lies in the interval:
$A \in \left(\frac{15}{16}, \frac{525}{256}\right)$

Step 4: Compare with the given options

Since:
$\frac{3}{4} < \frac{15}{16} \quad \text{and} \quad \frac{525}{256} < 3$

The interval $\left(\frac{15}{16}, \frac{525}{256}\right)$ is entirely contained within the interval $\left(\frac{3}{4}, 3\right)$.

Therefore, the area bounded by the curve lies in the interval $\left(\frac{3}{4}, 3\right)$.