Question

Find area bounded by region, y=3x+1, y=4x+1 and x=3?

Answer 1

The area of triangle ABC, bounded by the lines $y = 3x + 1, y = 2x + 1,$ and $x = 0,$ can be calculated as follows:

The area of triangle ABED is equal to the integral of the function (3x + 1) minus 0 with respect to x, evaluated from x = 0 to x = 4.

Similarly, the area of triangle ACED is equal to the integral of the function (2x + 1) minus 0 with respect to x, also evaluated from x = 0 to x = 4.

By subtracting the area of triangle ACED from the area of triangle ABED, we can find the area of triangle ABC.

Calculating the integrals:

$Area(ABED) = ∫[0 to 4] (3x + 1) dx - 0$

$Area(ACED) = ∫[0 to 4] (2x + 1) dx - 0$

Simplifying the integrals:

$Area(ABED) = ∫[0 to 4] 3x dx + ∫[0 to 4] 1 dx$

$Area(ACED) = ∫[0 to 4] 2x dx + ∫[0 to 4] 1 dx$

Evaluating the integrals:

$Area(ABED) = [3x^2 / 2] from 0 to 4 + [x] from 0 to 4$

$Area(ACED) = [2x^2 / 2] from 0 to 4 + [x] from 0 to 4$

Simplifying further:

$Area(ABED) = [(3 * 4^2) / 2] - [(3 * 0^2) / 2] + (4 - 0)$

$Area(ACED) = [(2 * 4^2) / 2] - [(2 * 0^2) / 2] + (4 - 0)$

Calculating:

$Area(ABED) = (48 / 2) - 0 + 4 = 24 + 4 = 28$ and

$Area(ACED) = (32 / 2) - 0 + 4 = 16 + 4 = 20$

Finally, we find the area of triangle ABC:

$Area(ABC) = Area(ABED) - Area(ACED) = 28 - 20 = 8$

Therefore, the area of triangle ABC is 8 square units.