wat the areb bounded by the curve y=3x^2-2x+1, the ordinates x=1 and x=3
Definite integral 1 to 3 ( 3 x ^ 2 - 2 x + 1 ) dx =
( x ^ 3 - x ^ 2 + x ) from 1 to 3 =
( 3 ^ 3 - 3 ^ 2 + 3 ) - ( 1 ^ 3 - 1 ^ 2 + 1 ) =
27 - 9 + 3 - ( 1 - 1 + 1 ) =
21 - 1 = 20
THANKS BOSNIAN
To find the area bounded by the curve and the ordinates x=1 and x=3, you can use definite integration.
First, let's graph the curve y=3x^2-2x+1 to visualize the region we are interested in.
To integrate the function, we will calculate the definite integral of y=3x^2-2x+1 between the limits x=1 and x=3.
The definite integral of a function f(x) between the limits a and b is given by:
∫(from a to b) f(x) dx
For this problem, we want to calculate the integral of y=3x^2-2x+1 between x=1 and x=3.
∫(from 1 to 3) (3x^2-2x+1) dx
To integrate, we use the power rule of integration.
The integral of x^n is (1/(n+1)) * x^(n+1), where n is not equal to -1.
Applying the power rule, we can integrate each term of the function separately.
∫(from 1 to 3) 3x^2 dx - ∫(from 1 to 3) 2x dx + ∫(from 1 to 3) 1 dx
Integrating each term, we get:
(3/3) * x^3 - (2/2) * x^2 + (1 * x)
Simplifying further:
x^3 - x^2 + x
Now, we substitute the limits (x=3 and x=1) into the integrated expression:
(3^3 - 3^2 + 3) - (1^3 - 1^2 + 1)
Simplifying:
(27 - 9 + 3) - (1 - 1 + 1)
Calculating:
21 - 1 = 20
Therefore, the area bounded by the curve y=3x^2-2x+1 and the ordinates x=1 and x=3 is 20 square units.