Find the definate integal (upper limit 1, lower limit 0) (2x^2-4x-6)dx
what is limit 2(x-3)(x+1)dx
Isn't this a integral uv dv problem?
Looks like a straight polynomial integral to me.
Integral (2x^2-4x-6)dx from 0 to 1
= (2/3)x^3 - 2x^2 - 6x from 0 to 1
= (2/3)(1)^3 - 2(1)^2 - 6(1) - ((2/3)(0)^3 - 2(0)^2 - 6(0))
= 2/3 - 2 - 6 - (0 - 0 - 0)
= 2/3 - 8
= 2/3 - 24/3
= -22/3
Hope this helps :3
To find the definite integral of the function (2x^2 - 4x - 6) from 0 to 1, you can follow these steps:
Step 1: Start with the given function: f(x) = 2x^2 - 4x - 6
Step 2: To find the definite integral, we need to find the antiderivative of the function. By applying the power rule for integration, we can find the antiderivative for each term of the function:
∫(2x^2 - 4x - 6) dx = (2/3)x^3 - 2x^2 - 6x + C
Step 3: Now, we will evaluate the antiderivative at the upper and lower limits of integration (1 and 0):
F(1) - F(0) = [(2/3)(1)^3 - 2(1)^2 - 6(1)] - [(2/3)(0)^3 - 2(0)^2 - 6(0)]
= [(2/3) - 2 - 6] - [0 - 0 - 0]
= (2/3) - 2 - 6
= 2/3 - 18/3 - 18/3
= (2 - 18 - 18)/3
= -34/3
Therefore, the value of the definite integral of (2x^2 - 4x - 6) from 0 to 1 is -34/3.