An Integral Problem by Lin Le 1

$\int_{0}^{1}x \; dx = \frac{1}{2}x^2 \bigg|_{0}^{1} = \frac{1}{2}(1)^2 - \frac{1}{2}(0)^2 = \green{\boxed{\frac{1}{2}}}$

The integral is asking for the area under $y=x$ between $x=0$ and $x=1$ . So if you graph $y=x$ between the bounds, you can find the area under the function by simply drawing a triangle with the hypotenuse as $y=x$ and finding the area of the triangle.

The length of the triangle would be 1 (since x goes from 0 to 1), and the height would be 1 (since y goes from 0 to 1).

Since the area of a triangle is given by $\frac{1}{2} bh$ , the area will be $\frac{1}{2} (1)(1)=\frac{1}{2}$ , or $0.5$ .