Question:

A function 𝑓(𝑥) is given by 𝑓(𝑥) = 𝑥^2 - 3𝑥 + 5.

(a) Find the indefinite integral of 𝑓(𝑥) with respect to 𝑥.

(b) Using the result from part (a), evaluate the definite integral of 𝑓(𝑥) from 𝑥 = 1 to 𝑥 = 3.

Answer:

(a) To find the indefinite integral of 𝑓(𝑥) with respect to 𝑥, we will use the power rule for integration. The power rule states that for any term 𝑥^𝑛, where 𝑛 is any real number except -1, the indefinite integral is given by:

∫𝑥^𝑛 𝑑𝑥 = (1/𝑛+1)𝑥^(𝑛+1) + 𝐶

where 𝐶 is the constant of integration.

Applying the power rule, we find:

∫(𝑥^2 - 3𝑥 + 5) 𝑑𝑥 = (1/3)𝑥^3 - (3/2)𝑥^2 + 5𝑥 + 𝐶

Therefore, the indefinite integral of 𝑓(𝑥) with respect to 𝑥 is:

∫𝑓(𝑥) 𝑑𝑥 = (1/3)𝑥^3 - (3/2)𝑥^2 + 5𝑥 + 𝐶

(b) Now, let's evaluate the definite integral of 𝑓(𝑥) from 𝑥 = 1 to 𝑥 = 3 using the result obtained in part (a).

∫[(1/3)𝑥^3 - (3/2)𝑥^2 + 5𝑥] 𝑑𝑥 evaluated from 𝑥 = 1 to 𝑥 = 3

Using the fundamental theorem of calculus, we can evaluate the definite integral by subtracting the value of the antiderivative at the upper limit from the value of the antiderivative at the lower limit.

Let's substitute 𝑥 = 3 into the antiderivative:

(1/3)(3)^3 - (3/2)(3)^2 + 5(3) = 9 - 13.5 + 15 = 10.5

Next, let's substitute 𝑥 = 1 into the antiderivative:

(1/3)(1)^3 - (3/2)(1)^2 + 5(1) = 1/3 - 3/2 + 5 = -3/2 + 5 + 1/3 = 1/2 + 5 + 1/3 = 6 + 1/2 + 1/3 = 6.833

Finally, we subtract the values at the upper limit and lower limit:

10.5 - 6.833 = 3.667

Therefore, the value of the definite integral of 𝑓(𝑥) from 𝑥 = 1 to 𝑥 = 3 is approximately 3.667.