Q.Find the area of the region bounded by the line $y = 3x + 1$, the $x$-axis, and the ordinates $x = 1$ and $x = 3$.

The triangle bounded by $y=4x$, the x-axis, and $x=4$ is found by integrating the line's height from $x=0$ to $x=4$.

The line $y=4x$ meets the x-axis at the origin $(0,0)$, and together with the vertical line $x=4$, it bounds a right triangle with vertices $(0,0)$, $(4,0)$, and $(4,16)$.

Using calculus, the area under $y=4x$ from $x=0$ to $x=4$ is:

$A = \int_0^4 4x\,dx = 4\left[\frac{x^2}{2}\right]_0^4 = 2(16-0) = 32$

(Check with the geometric formula: $\dfrac12\times\text{base}\times\text{height} = \dfrac12\times4\times16=32$ — matches.)

The region under the line $y=x$ from $x=0$ to $x=2$ is found by the definite integral $\int_0^2 x\,dx$.

Step 1. The line $y=x$, the vertical line $x=2$, and the $x$-axis bound a right triangle with vertices $(0,0),(2,0),(2,2)$.

Step 2. By calculus, Area $=\displaystyle\int_0^2 y\,dx = \int_0^2 x\,dx = \left[\frac{x^2}{2}\right]_0^2 = \frac{4}{2}-0 = 2$.

Step 3. (Check: this matches the elementary triangle-area formula $\tfrac12\times\text{base}\times\text{height}=\tfrac12\times2\times2=2$.)