Find the definate integal (upper limit 1, lower limit 0) (7-4x)^2dx
let u=(7-4x)
then du=4dx
so the integration is now..
INT u^2 du/4 limits from 7 to 3
looks like u^3/12 limits 7 to 3
check that
To find the definite integral of the function (7-4x)^2 with upper limit 1 and lower limit 0, follow these steps:
Step 1: Expand the expression (7-4x)^2 using the binomial formula or by squaring the binomial.
(7-4x)^2 = (7-4x)(7-4x) = 49 - 28x + 16x^2
Step 2: Integrate each term of the expanded expression. The integral of a constant is the constant times the variable, while the integral of x raised to a power is (1/n+1)x^(n+1), where n is the power.
∫ (49 - 28x + 16x^2) dx = 49x - 28/2 x^2 + 16/3 x^3 + C
Step 3: Evaluate the definite integral by substituting the upper and lower limits into the antiderivative.
∫(0 to 1) (49 - 28x + 16x^2) dx = [49(1) - 28/2(1)^2 + 16/3(1)^3] - [49(0) - 28/2(0)^2 + 16/3(0)^3]
Simplifying further:
∫(0 to 1) (49 - 28x + 16x^2) dx = [49 - 14 + 16/3] = 21.33
Therefore, the definite integral of (7-4x)^2 with upper limit 1 and lower limit 0 is approximately equal to 21.33.