Question

Find the area under the graph of f (x)=2x^2-3x+2
and the x-axis x = 0 and x = 3

Answer 1

ANSWER:

Given curve is x^2 −3x+2=0

The roots of this quadratic equation are 1 and2

Thus, the area under the curve and X-axis and the ordinates x=0 x=3 is

A=∫ 03 (x ^2 −3x+2) A=∫ 01 (x ^2−3x+2)−∫ 12(x^2−3x+2)+∫ 23(x 2−3x+2) ....... (Between x=1 to x=2, the curve is below X-axis)

After integrating and substituting upper and lower limits, we get

A= 6÷5 + 6÷1 + 6÷5

∴A= 6÷11

Hope you can relate

Plz give me 5 star

Answer 2

Answer:

The area in between the curve  $f (x)=2x^2-3x+2$  and line $y=0$ with limits of x between 0 and 3 can be given by

$A=\int\limits^3_0 {(2x^2-3x+2)} \, dx$

   $=2\int\limits^3_0 {x^2} \, dx -3\int\limits^3_0 {x} \, dx+\int\limits^3_0 {2} \, dx$

   $=|\frac{2x^3}{3} -\frac{3x^2}{2} +2x|^3_0$

   $=\frac{2(3)^3}{3} -\frac{3(3)^2}{2} +2(3)-\frac{2(0)^3}{3} +\frac{3(0)^2}{2} +2(0)$

    $=18-\frac{27}{2} +6-0+0-0$

   $=\frac{48-27}{2} \\=\frac{21}{2} =10.5$

∴ Area under the curve  $f (x)=2x^2-3x+2$   is  10.5 square units.