The curve $C$ passes through the points with coordinates $(a, 0)$, $(-1, 0)$, $(b, 0)$, and $(0, c)$, where $a < -1$, $b > 0$, $c > 0$.
Given that $f'(x) = 17 + 2x - 3x^2$:
(a) Show that the equation of $C$ is $y = 15 + 17x + x^2 - x^3$
To find $f(x)$, integrate $f'(x)$:
$ f(x) = \int \left(17 + 2x - 3x^2 \right) \, dx. $
$ f(x) = 17x + x^2 - x^3 + C, $
where $C$ is the constant of integration.
Since $C$ passes through the point $(-1, 0)$, substitute $(-1, 0)$ into $f(x)$:
$ 0 = 17(-1) + (-1)^2 - (-1)^3 + C. $
$ 0 = -17 + 1 + 1 + C. $
$ C = 15. $
Thus, the equation of $C$ is:
$ y = f(x) = 15 + 17x + x^2 - x^3. $
(b) Find the values of $a$, $b$, and $c$
The roots of $f(x) = 0$ give the $x$-intercepts $(a, 0)$, $(-1, 0)$, and $(b, 0)$. Solve:
$ 15 + 17x + x^2 - x^3 = 0. $
Factorize:
$ -x^3 + x^2 + 17x + 15 = 0 \implies -(x+1)(x^2 - 2x - 15) = 0. $
The factor $x+1 = 0$ gives $x = -1$. Solve $x^2 - 2x - 15 = 0$ using the quadratic formula:
$ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)} = \frac{2 \pm \sqrt{4 + 60}}{2}. $
$ x = \frac{2 \pm \sqrt{64}}{2} = \frac{2 \pm 8}{2}. $
Thus:
$ x = \frac{2 + 8}{2} = 5, \quad x = \frac{2 - 8}{2} = -3. $
So, $a = -3$, $b = 5$.
To find $c$, substitute $x = 0$ into $f(x)$:
$ c = f(0) = 15 + 17(0) + (0)^2 - (0)^3 = 15. $
(c) Use algebraic integration to find the exact area of region $R$
The region $R$ is bounded by the curve $C$ and the straight line $l$, which passes through the points $(b, 0)$ and $(0, c)$. The equation of $l$ is:
$ y = mx + c, $
where $m$ is the slope. The slope is:
$ m = \frac{c - 0}{0 - b} = \frac{15}{-5} = -3. $
Thus, the equation of $l$ is:
$ y = -3x + 15. $
The area of $R$ is given by:
$ \text{Area} = \int_0^5 \left(f(x) - l(x)\right) \, dx. $
Substitute $f(x)$ and $l(x)$:
$ \text{Area} = \int_0^5 \left(15 + 17x + x^2 - x^3 - (-3x + 15)\right) dx. $
$ \text{Area} = \int_0^5 \left(-x^3 + x^2 + 20x \right) dx. $
Simplify the integration term by term:
$ \int_0^5 -x^3 \, dx = \left[-\frac{x^4}{4}\right]_0^5 = -\frac{5^4}{4} = -\frac{625}{4}, $
$ \int_0^5 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^5 = \frac{5^3}{3} = \frac{125}{3}, $
$ \int_0^5 20x \, dx = \left[10x^2\right]_0^5 = 10(5^2) = 250. $
Combine the results:
$ \text{Area} = -\frac{625}{4} + \frac{125}{3} + 250. $
Find the common denominator and simplify:
$ \text{Area} = \frac{-1875 + 500 + 3000}{12} = \frac{1625}{12}=135\frac{5}{12}. $
Thus, the exact area of $R$ is:
$ \boxed{135\frac{5}{12}}. $
Final Answers:
- (a) The equation of $C$ is:
$ y = 15 + 17x + x^2 - x^3. $
- (b) $a = -3$, $b = 5$, $c = 15$.
- (c) The exact area of $R$ is:
$ 135\frac{5}{12}. $
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