We are asked to find the area bounded by the function $f(x) = \frac{4}{3}x^3 - 7.5x^2 + 10x$, the x-axis, and the line $x = C$, where $C$ is the x-coordinate of the maximum point of the function $f(x)$. We need to find the value of $C$, and then calculate the definite integral of $f(x)$ from 0 to $C$.

1. Problem Description

We are asked to find the area bounded by the function $f(x) = \frac{4}{3}x^3 - 7.5x^2 + 10x$, the x-axis, and the line $x = C$, where $C$ is the x-coordinate of the maximum point of the function $f(x)$. We need to find the value of $C$, and then calculate the definite integral of $f(x)$ from 0 to $C$.

2. Solution Steps

First, we need to find the critical points of the function $f(x)$. To do this, we find the first derivative and set it equal to zero.

$f(x) = \frac{4}{3}x^3 - 7.5x^2 + 10x$

$f'(x) = 4x^2 - 15x + 10$

Now we set $f'(x) = 0$ to find the critical points.

$4x^2 - 15x + 10 = 0$

We can use the quadratic formula to solve for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{15 \pm \sqrt{(-15)^2 - 4(4)(10)}}{2(4)}$

$x = \frac{15 \pm \sqrt{225 - 160}}{8}$

$x = \frac{15 \pm \sqrt{65}}{8}$

$x_1 = \frac{15 + \sqrt{65}}{8} \approx 3.133$

$x_2 = \frac{15 - \sqrt{65}}{8} \approx 0.617$

To determine which critical point corresponds to the maximum, we can find the second derivative of $f(x)$:

$f''(x) = 8x - 15$

Now we evaluate the second derivative at each critical point:

$f''(3.133) = 8(3.133) - 15 = 25.064 - 15 = 10.064 > 0$, so $x_1 \approx 3.133$ is a local minimum.

$f''(0.617) = 8(0.617) - 15 = 4.936 - 15 = -10.064 < 0$, so $x_2 \approx 0.617$ is a local maximum.

Therefore, $C = \frac{15 - \sqrt{65}}{8} \approx 0.617$.

Now we need to find the area under the curve from $x = 0$ to $x = C$:

$Area = \int_0^C f(x) dx = \int_0^{\frac{15 - \sqrt{65}}{8}} (\frac{4}{3}x^3 - 7.5x^2 + 10x) dx$

$Area = [\frac{1}{3}x^4 - 2.5x^3 + 5x^2]_0^{\frac{15 - \sqrt{65}}{8}}$

Let $C = \frac{15 - \sqrt{65}}{8}$.

$Area = \frac{1}{3}C^4 - 2.5C^3 + 5C^2$

Using $C \approx 0.617$:

$Area = \frac{1}{3}(0.617)^4 - 2.5(0.617)^3 + 5(0.617)^2$

$Area = \frac{1}{3}(0.145) - 2.5(0.235) + 5(0.381)$

$Area = 0.048 - 0.588 + 1.905$

$Area = 1.365$

3. Final Answer

The area is approximately 1.

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