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

Nov 15, 2016

1 square unit

#### Explanation:

${\int}_{0}^{1} \left(2 x\right) \mathrm{dx}$

You can solve this many ways:

1. (easiest)Think of it geometrically, as a line $y = 2 x$, 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 $f \left(x\right) = 2 x$, and $F \left(x\right)$ is the antiderivative of $f \left(x\right)$, then ${\int}_{0}^{1} \left(2 x\right) \mathrm{dx} = F \left(1\right) - F \left(0\right)$

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:
${\int}_{0}^{1} \left(2 x\right) \mathrm{dx}$
Let $f \left(x\right) = 2 x$
$F \left(x\right) = \int 2 x$
$= {x}^{2} + c$
$F \left(1\right) - F \left(0\right)$
$= 1 - 0$
$= 1$square unit