+130516 Have a look at the graph here.........https://www.desmos.com/calculator/wk4fnvhslz
The area bounded by x= 1, x = 2 and the x axis is almost a triangle with a base of 1 and a height of about 7/ 10
So the area [ estimated] is about (1/2)(1)(7/10) = 7/20 = 0.35 sq units
BTW ....The actual value is about 0.38629 sq units
+26400 estimate the area under the curve y(x) = ln(x) from x=1 to x=2
\(\boxed{~ \int \limits_{x=1}^{x=2} \ln{(x)}\ dx = \ ? ~ }\)
\(\text{Integration by parts } \boxed{~ \int u'v= uv - \int uv' ~} \\ \begin{array}{rclcrcl} u' &=& 1 && v&=& \ln{(x)} \\ u &=& x && v' &=& \frac{1}{x}\\\\ \int 1\cdot \ln{(x)} \ dx &=& x\cdot \ln{(x)} -\int x\cdot \frac{1}{x} \ dx\\ &=& x\cdot \ln{(x)} -\int \ dx\\ \int 1\cdot \ln{(x)} \ dx &=& x\cdot \ln{(x)} - x\\\\ \end{array}\\ \begin{array}{rcl} \int \limits_{x=1}^{x=2} \ln{(x)}\ dx &=& \left[ x\cdot \ln{(x)} - x \right]_{1}^{2}=2\cdot \ln{(2)-2} - [ 1\cdot \ln{(1)}-1] \qquad | \qquad \ln{(1)} = 0 \\ &=& 2\cdot \ln{(2)}-2 - (-1)\\ &=& 2\cdot \ln{(2)}-2 +1\\ &=& 2\cdot \ln{(2)}-1\\ \mathbf{ \int \limits_{x=1}^{x=2} \ln{(x)}\ dx }& \mathbf{=} & \mathbf{ 0.38629436112 } \end{array}\)
The area under the curve y(x) = ln(x) from x=1 to x=2 is 0.38629436112
