Use the limit definition of definite integral to evaluate \displaystyle{ \int^{4}_{0} 5 \, dx } .
L Lizz New member Joined Nov 24, 2009 Messages 1 Nov 25, 2009 #1 Use the limit definition of definite integral to evaluate ∫045 dx\displaystyle \displaystyle{ \int^{4}_{0} 5 \, dx }∫045dx .
Use the limit definition of definite integral to evaluate ∫045 dx\displaystyle \displaystyle{ \int^{4}_{0} 5 \, dx }∫045dx .
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Nov 25, 2009 #2 I reckon you mean a Riemann sum?. If we calculate the area of an infinite number of rectangles, we approach the area under the curve, such as it is. Right endpoint method: a+kΔx\displaystyle a+k{\Delta}xa+kΔx Δx=4−0n\displaystyle {\Delta}x=\frac{4-0}{n}Δx=n4−0 Thus, rectangle k has area: f(xk)Δx=5Δx=54n=20n\displaystyle f(x_{k}){\Delta}x=5{\Delta}x=5\frac{4}{n}=\frac{20}{n}f(xk)Δx=5Δx=5n4=n20 The sum of the areas is ∑k=1n20n=1n∑k=1n20=20\displaystyle \sum_{k=1}^{n}\frac{20}{n}=\frac{1}{n}\sum_{k=1}^{n}20=20k=1∑nn20=n1k=1∑n20=20
I reckon you mean a Riemann sum?. If we calculate the area of an infinite number of rectangles, we approach the area under the curve, such as it is. Right endpoint method: a+kΔx\displaystyle a+k{\Delta}xa+kΔx Δx=4−0n\displaystyle {\Delta}x=\frac{4-0}{n}Δx=n4−0 Thus, rectangle k has area: f(xk)Δx=5Δx=54n=20n\displaystyle f(x_{k}){\Delta}x=5{\Delta}x=5\frac{4}{n}=\frac{20}{n}f(xk)Δx=5Δx=5n4=n20 The sum of the areas is ∑k=1n20n=1n∑k=1n20=20\displaystyle \sum_{k=1}^{n}\frac{20}{n}=\frac{1}{n}\sum_{k=1}^{n}20=20k=1∑nn20=n1k=1∑n20=20
