Question

Using the method of integration show that the area of triangle of base $b$ and altitude $h$ is $\frac{1}{2}bh$.

Answer 1

Let $ABC$ be the triangle with base $BC=b$ and altitude $AM=h$
Let us consider a thin strip $DE$ on the triangle of thickness $dx$, at a distance $x$ from the vertex $A$, parallel to the base $BC$.
If $y$ be the length of the strip $DE$, then from similar triangles $ABC$ and $ADE$, we have $\frac{y}{x}=\frac{b}{h}$.
$\Rightarrow y=\frac{bx}{h}$
Therefore, the area of the rectangular strip $DE$ is given by
$dA=ydx=\frac{bxdx}{h}$
The complete area of the triangle can be obtained by summing up (integration) the area of individual strips such as $DE$.
$A=\sum dA={\int }_0^h\frac{bxdx}{h}=\frac{b}{h}{\left[\frac{x^2}{2}\right]}^h=\frac{b}{2h}(h^2-0)=\frac{1}{2}bh$.