6FExample1

Finding Area Example 1

Find, correct to 2 decimal places, the area bounded by math . \quad \cdot \quad \text{and the curve } f(x)=3xe^{-x^2} math
 * **x-axis**,
 * the line **x = –1**,
 * the line **x = 2**

First: Sketch and observe that the area is split above and below the x-axis. The area will have to be calculated in 2 sections.

From observation of the graph, or by solving **//f(x)=0//** , we find the x-intercept is at **x = 0**.

math . \qquad \text{Area } \; = -\displaystyle{\int\limits_{x=-1}^{x=0} 3xe^{-x^2} } \; dx + \displaystyle{\int\limits_{x=0}^{x=2} 3xe^{-x^2} } \; dx math

math . \qquad \displaystyle{\int 3xe^{-x^2} }dx \quad \text{can be evaluated using substitution method:} math . math . \qquad \text{Let } \; u=x^2 \; \Rightarrow \; dx= \dfrac{du}{2x} math . math . \qquad \displaystyle{\int 3xe^{-x^2} }dx \; = \displaystyle{\int 3xe^{-u} }\dfrac {du}{2x} math . math \\ . \qquad \qquad \qquad \; = \frac{3}{2} \int{e^{-u} } \; du \\. \\ . \qquad \qquad \qquad \; = - \frac{3}{2} e^{-u} + c \\. \\ . \qquad \qquad \qquad \; = - \frac{3}{2} e^{-x^2} + c math . Thus math . \qquad \text{Area } \; = -\displaystyle{\int\limits_{x=-1}^{x=0} 3xe^{-x^2} } \; dx + \displaystyle{\int\limits_{x=0}^{x=2} 3xe^{-x^2} } \; dx math . math . \qquad \qquad = - \left[ - \dfrac{3}{2} e^{-x^2} \right]_{x=-1}^{x=0} + \left[ - \dfrac{3}{2} e^{-x^2} \right]_{x=0}^{x=2} math . math . \qquad \qquad = - \left[ - \frac{3}{2} e^{0} - \left( - \frac{3}{2} e^{-1} \right) \right] + \left[ - \frac{3}{2} e^{-4} - \left( - \frac{3}{2} e^{0} \right) \right] math . math . \qquad \qquad = - \left[ - \frac{3}{2} + \frac{3}{2} e^{-1} \right] + \left[ - \frac{3}{2} e^{-4} + \frac{3}{2} \right] math . math . \qquad \qquad = \frac{3}{2} - \frac{3}{2} e^{-1} - \frac{3}{2} e^{-4} + \frac{3}{2} math . math . \qquad \qquad = 3 - \frac{3}{2} e^{-1} - \frac{3}{2} e^{-4} \; \text{ square units} math . math . \qquad \qquad = 2.42 \; \text{ square units} math

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