calculate the area bounded by the graphs of f(x)=x2 , g(x)=1/x2, x>0 and the line x=3 :)
+26387 calculate the area bounded by the graphs of f(x)=x2 , g(x)=1/x2, x>0 and the line x=3 :)
1. cut\( f(x) \text{ and } g(x) , x \gt 0\)
\(\begin{array}{rcll} f(x) &=& g(x)\\ x^2 &=& \frac{1}{x^2} \\ x^4 &=& 1 \\ x &=& 1 \end{array}\)
2. calculate the area
\(\begin{array}{rcll} \int \limits_{1}^{3} { ( f(x)-g(x) )\ dx } &=& \int \limits_{1}^{3} { f(x) \ dx } -\int \limits_{1}^{3} { g(x) \ dx } \\ &=& \int \limits_{1}^{3} { x^2 \ dx } -\int \limits_{1}^{3} { \frac{1}{x^2} \ dx } \\ &=& [\frac{x^3}{3}]_{1}^{3} - [-\frac{1}{x} ]_{1}^{3} \\ &=& \frac13\cdot [x^3]_{1}^{3} + [ \frac{1}{x} ]_{1}^{3} \\ &=& \frac13\cdot (3^3-1^3) + ( \frac{1}{3}-\frac{1}{1} )\\ &=& \frac13\cdot (27-1) + ( -\frac{2}{3} )\\ &=& \frac13\cdot (26) -\frac{2}{3} \\ &=& \frac{26-2}{3} \\ &=& \frac{24}{3} \\ &=& \frac{3\cdot 8}{3} \\ \mathbf{ \int \limits_{1}^{3} { ( f(x)-g(x) )\ dx } } & \mathbf{=} & \mathbf{8}\\ \end{array}\)
+26387 calculate the area bounded by the graphs of f(x)=x2 , g(x)=1/x2, x>0 and the line x=3 :)
1. cut\( f(x) \text{ and } g(x) , x \gt 0\)
\(\begin{array}{rcll} f(x) &=& g(x)\\ x^2 &=& \frac{1}{x^2} \\ x^4 &=& 1 \\ x &=& 1 \end{array}\)
2. calculate the area
\(\begin{array}{rcll} \int \limits_{1}^{3} { ( f(x)-g(x) )\ dx } &=& \int \limits_{1}^{3} { f(x) \ dx } -\int \limits_{1}^{3} { g(x) \ dx } \\ &=& \int \limits_{1}^{3} { x^2 \ dx } -\int \limits_{1}^{3} { \frac{1}{x^2} \ dx } \\ &=& [\frac{x^3}{3}]_{1}^{3} - [-\frac{1}{x} ]_{1}^{3} \\ &=& \frac13\cdot [x^3]_{1}^{3} + [ \frac{1}{x} ]_{1}^{3} \\ &=& \frac13\cdot (3^3-1^3) + ( \frac{1}{3}-\frac{1}{1} )\\ &=& \frac13\cdot (27-1) + ( -\frac{2}{3} )\\ &=& \frac13\cdot (26) -\frac{2}{3} \\ &=& \frac{26-2}{3} \\ &=& \frac{24}{3} \\ &=& \frac{3\cdot 8}{3} \\ \mathbf{ \int \limits_{1}^{3} { ( f(x)-g(x) )\ dx } } & \mathbf{=} & \mathbf{8}\\ \end{array}\)
