How do you evaluate the definite integral #int2xdx# from #[0,1]#?

1 Answer
Anjali G
Nov 15, 2016

1 square unit

Explanation:

#int_0^1 (2x)dx#

You can solve this many ways:

  1. (easiest)Think of it geometrically, as a line #y=2x#, and find the area under the curve from #0# to #1# using the area of a triangle formula

  2. (easy)Use a fundamental theorem of calculus, and say that if #f(x)=2x#, and #F(x)# is the antiderivative of #f(x)#, then #int_0^1 (2x)dx=F(1)-F(0)#

  3. (hard)Use Riemann sums to find the area under the curve from #0# to #1# using an infinite number of rectangles

Method #1: graph{2x [-2.784, 2.69, -0.523, 2.215]} Area of triangle#=(1/2)(b)(h)# #=(1/2)(1)(2)# #=1# square unit

Method 2:
#int_0^1 (2x)dx#
Let #f(x)=2x#
#F(x)=int2x#
#=x^2+c#
#F(1)-F(0)#
#=1-0#
#=1 #square unit

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