Evaluate integral upper limit 1 and lower limit 0 of (7x^2 - 8x +3)dx

To evaluate the integral ∫[0 to 1] (7x^2 - 8x + 3) dx, you can use the rules of integration.

Step 1: Find the antiderivative of each term.

The antiderivative of 7x^2 is (7/3)x^3.
The antiderivative of -8x is -4x^2.
The antiderivative of 3 is 3x.

Step 2: Evaluate the antiderivatives at the upper limit and the lower limit.

At x = 1,
(7/3)(1)^3 - 4(1)^2 + 3(1) = 7/3 - 4 + 3 = 7/3 - 4 + 3 = 7/3 - 4/1 + 3/1 = 7/3 - 12/3 + 9/3 = 4/3.

At x = 0,
(7/3)(0)^3 - 4(0)^2 + 3(0) = 0 - 0 + 0 = 0.

Step 3: Subtract the lower limit value from the upper limit value.

4/3 - 0 = 4/3.

Therefore, the value of the given integral is 4/3.