Evaluate Integral upper limit 2 and lower limit 0 of 3x(x-7)dx
To evaluate the integral ∫(0 to 2) 3x(x-7) dx, we can use the integral rules to simplify the expression and then apply the power rule for integration.
First, let's expand the expression inside the integral:
∫(0 to 2) 3x(x-7) dx = ∫(0 to 2) 3x^2 - 21x dx.
Next, we can integrate each term separately using the power rule. The power rule states that for any term of the form x^n, the integral is (1/(n+1)) * x^(n+1).
Integrating the first term, 3x^2, we add 1 to the exponent and divide by the new exponent:
∫(0 to 2) 3x^2 dx = (1/3) * x^3 |(0 to 2) = (1/3) * (2^3 - 0^3) = (1/3) * 8 = 8/3.
Integrating the second term, -21x, we add 1 to the exponent and divide by the new exponent:
∫(0 to 2) -21x dx = (-21/2) * x^2 |(0 to 2) = (-21/2) * (2^2 - 0^2) = (-21/2) * 4 = -42.
Now, we can combine the results of both integrals:
∫(0 to 2) 3x(x-7) dx = 8/3 - 42 = 8/3 - 42/1 = 8/3 - 126/6 = (8 - 126)/6 = -118/6 = -59/3.
Therefore, the value of the integral ∫(0 to 2) 3x(x-7) dx is -59/3.