Q.Find the area of the region bounded by the line $y = 3x$, the $y$-axis, and the line $y = 6$.

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: …

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

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