Question

Find the definite integral of the function f(x) = 3x^2 - 2x + 1 from x = 1 to x = 4.

Answer

To find the definite integral of the function f(x) = 3x^2 - 2x + 1 from x = 1 to x = 4, we will use the definite integral formula.

The definite integral of a function f(x) over the interval [a, b] is given by:

∫[a to b] f(x) dx

So, in this case, we have:

∫[1 to 4] (3x^2 - 2x + 1) dx

To solve this, we will find the antiderivative of the function and then evaluate it from x = 1 to x = 4.

The antiderivative of 3x^2 is x^3, the antiderivative of -2x is -x^2, and the antiderivative of 1 is x.

So, using the antiderivative, we have:

x^3 - x^2 + x |[1,4]

Now we can evaluate the antiderivative at the upper and lower limits:

= (4^3 - 4^2 + 4) - (1^3 - 1^2 + 1) = (64 - 16 + 4) - (1 - 1 + 1) = 52 - 1 = 51

Therefore, the definite integral of the function f(x) = 3x^2 - 2x + 1 from x = 1 to x = 4 is 51.

So, ∫[1 to 4] (3x^2 - 2x + 1) dx = 51.