Question

find the area bounded by y=x^3+3x^2 and y=2x^2+4x

Answer 1

find the area bounded by $y = x^3 + 3x^2$ and $y = 2x^2 + 4x$

If we must to find the area $S$ that is bounded by $y_{1}(x)$ and $y_{2}(x)$ we use the formula:

$S = \int_{x_1}^{x_2}(y_2(x) - y_1(x))dx,$ where $x_{1}$ and $x_{2}$ are the point of intersection of $y_{2}(x)$ and $y_{1}(x)$

$y_{1}(x) = x^{3} + 3x^{2}$

$y_{2}(x) = 2x^{2} + 4x$

$x^{3} + 3x^{2} = 2x^{2} + 4x$

$x_{1} = \frac{1}{2} (-1 - \sqrt{17})$

$x_{2} = \frac{1}{2} (-1 + \sqrt{17})$

$x_{3} = 0$

$S = \int_{\frac{1}{2}(-1 - \sqrt{17})}^{0}(x^{3} + 3x^{2} - 2x^{2} - 4x)dx + \int_{0}^{\frac{1}{2}(-1 + \sqrt{17})}(2x^{2} + 4x - x^{3} - 3x^{2})dx =$

$\int_{\frac{1}{2}(-1 - \sqrt{17})}^{0}(x^{3} + x^{2} - 4x)dx + \int_{0}^{\frac{1}{2}(-1 + \sqrt{17})}(4x - x^{3} - x^{2})dx =$

$$= \left(\frac {x ^ {4}}{4} + \frac {x ^ {3}}{3} - \frac {4 x ^ {2}}{2}\right) \left| _ {\frac {1}{2} (- 1 - \sqrt {1 7})} ^ {0} + \left(\frac {4 x ^ {2}}{2} - \frac {x ^ {4}}{4} - \frac {x ^ {3}}{3}\right) \right| _ {0} ^ {\frac {1}{2} (- 1 + \sqrt {1 7})} = \frac {1 2 1}{1 2}$$