Question

Consider the function f(x) and g(x) defined from

$\mathbb{R} \to \mathbb{R}, f(x)=\frac{x^{3}}{2}+1-x\int_{0}^{x}g(t)dt, g(x)=x-\int_{0}^{1}f(t)dt$, then area bounded by $f(x), g(x), x=0$ and $x=2, y \ge 0$ is $\frac{p}{q}$ (p and q are coprimes), then (p + q) is equal to _____.

Answer 1

55

Answer 2

Let $C = \int_{0}^{1}f(t)dt$. Since the integral is from a constant limit to a constant limit, $C$ is a constant. From the definition of $g(x)$, we have: $g(x) = x - C$

Substitute this expression for $g(x)$ into the equation for $f(x)$: $f(x)=\frac{x^{3}}{2}+1-x\int_{0}^{x}(t-C)dt$

Evaluate the integral: $\int_{0}^{x}(t-C)dt = \left[\frac{t^2}{2} - Ct\right]_{0}^{x} = \frac{x^2}{2} - Cx$

Substitute this back into the equation for $f(x)$: $f(x)=\frac{x^{3}}{2}+1-x\left(\frac{x^2}{2} - Cx\right)$ $f(x)=\frac{x^{3}}{2}+1-\frac{x^3}{2} + Cx^2$ $f(x) = Cx^2 + 1$

Now, use the definition of $C$ to find its value: $C = \int_{0}^{1}f(t)dt$ Substitute $f(t) = Ct^2 + 1$: $C = \int_{0}^{1}(Ct^2 + 1)dt$ $C = \left[C\frac{t^3}{3} + t\right]_{0}^{1}$ $C = \left(C\frac{1^3}{3} + 1\right) - (0+0)$ $C = \frac{C}{3} + 1$

Solve for $C$: $C - \frac{C}{3} = 1$ $\frac{2C}{3} = 1$ $C = \frac{3}{2}$

So, the explicit forms of the functions are: $f(x) = \frac{3}{2}x^2 + 1$ $g(x) = x - \frac{3}{2}$

We need to find the area bounded by $f(x), g(x), x=0$ and $x=2$, with the condition $y \ge 0$. This means we are looking for the area of the region $R = \{(x, y) \mid 0 \le x \le 2, \max(0, g(x)) \le y \le f(x)\}$.

First, let's determine the relationship between $f(x)$ and $g(x)$. Consider $f(x) - g(x)$: $f(x) - g(x) = (\frac{3}{2}x^2 + 1) - (x - \frac{3}{2}) = \frac{3}{2}x^2 - x + \frac{5}{2}$ The discriminant of $3x^2 - 2x + 5$ is $\Delta = (-2)^2 - 4(3)(5) = 4 - 60 = -56 < 0$. Since the leading coefficient is positive, $f(x) - g(x)$ is always positive, meaning $f(x) > g(x)$ for all $x$.

Next, find where $g(x)$ is positive: $g(x) = x - \frac{3}{2} > 0 \implies x > \frac{3}{2}$. So, $g(x)$ is positive for $x \in (\frac{3}{2}, 2]$ and negative for $x \in [0, \frac{3}{2})$.

The area is given by the integral: Area $= \int_{0}^{2} (f(x) - \max(0, g(x))) dx$

Split the integral based on the sign of $g(x)$: Area $= \int_{0}^{3/2} (f(x) - 0) dx + \int_{3/2}^{2} (f(x) - g(x)) dx$

Calculate the first integral: $\int_{0}^{3/2} f(x) dx = \int_{0}^{3/2} (\frac{3}{2}x^2 + 1) dx = \left[\frac{x^3}{2} + x\right]_{0}^{3/2} = \frac{(3/2)^3}{2} + \frac{3}{2} = \frac{27/8}{2} + \frac{3}{2} = \frac{27}{16} + \frac{24}{16} = \frac{51}{16}$

Calculate the second integral: $\int_{3/2}^{2} (f(x) - g(x)) dx = \int_{3/2}^{2} \left(\frac{3}{2}x^2 - x + \frac{5}{2}\right) dx = \left[\frac{x^3}{2} - \frac{x^2}{2} + \frac{5}{2}x\right]_{3/2}^{2}$ At $x=2$: $\frac{2^3}{2} - \frac{2^2}{2} + \frac{5}{2}(2) = 4 - 2 + 5 = 7$ At $x=3/2$: $\frac{(3/2)^3}{2} - \frac{(3/2)^2}{2} + \frac{5}{2}(3/2) = \frac{27}{16} - \frac{9}{8} + \frac{15}{4} = \frac{27 - 18 + 60}{16} = \frac{69}{16}$ So, the second integral is: $7 - \frac{69}{16} = \frac{112}{16} - \frac{69}{16} = \frac{43}{16}$

Total Area = $\frac{51}{16} + \frac{43}{16} = \frac{94}{16} = \frac{47}{8}$

The area is given as $\frac{p}{q}$, where $p$ and $q$ are coprimes. Here, $p=47$ and $q=8$. They are coprime. We need to find $p+q$: $p+q = 47 + 8 = 55$.