Evaluate the definite integral of 1 with respect to x from 0 to 2 pi.

Step-by-step solution

1. 1

   Interpret the integral geometrically first. The graph of $y=1$ is a horizontal line at height $1$, so the integral is the area of a rectangle of height $1$ over the interval $[0,2\pi]$. Its area is base times height, i.e. the length of the interval — so the answer should be $2\pi$ before any calculus is done.

2. 2

   Find an antiderivative. We need $F$ with $F'(x)=1$, and

   $\frac{d}{dx}(x)=1$

   so $F(x)=x$ works. (This is the power rule with $n=0$: $\int x^0dx=\tfrac{x^1}{1}$.)

3. 3

   Apply the Fundamental Theorem of Calculus.

   $\int_{0}^{2\pi}1\,dx=\Bigl[x\Bigr]_{0}^{2\pi}=2\pi-0=2\pi$

4. 4

   Give the numerical value.

   $2\pi=6.283185307\ldots\approx6.2832$

5. 5

   Note the general rule this illustrates. For any constant $c$ and limits $a<b$,

   $\int_a^b c\,dx=c(b-a)$

   So an integrand of $1$ measures length, which is exactly why $\int_a^b1\,dx$ appears as the "total time" or "total mass" term in applied integrals — and why the limit $2\pi$ here, one full revolution, shows up constantly in trigonometric work.

6. 6

   Cross-check the rectangle reading. Height $1$ times base $2\pi-0=2\pi$ gives $2\pi$ ✓, matching the Fundamental Theorem result exactly.