Evaluate the definite integral ∫ from 0 to 4 of (3x + 2) dx.

Evaluate the definite integral ∫ from 0 to 4 of (3x + 2) dx.

• A. 32
• B. 40
• C. 48
• D. 24

Answer: (B)

To evaluate the definite integral of the function \(3x + 2\) from 0 to 4, we can follow the process of finding the antiderivative of the function first. The function \(3x + 2\) can be integrated using the power rule. The antiderivative of \(3x\) is \(\frac{3}{2}x^2\), and the antiderivative of the constant \(2\) is \(2x\). Therefore, the combined antiderivative of the function \(3x + 2\) is: \[ F(x) = \frac{3}{2}x^2 + 2x \] Next, we will evaluate \(F(x)\) at the bounds \(0\) and \(4\), and then subtract the value at the lower limit from the value at the upper limit: 1. Calculate \(F(4)\): \[ F(4) = \frac{3}{2}(4^2) + 2(4) = \frac{3}{2}(16) + 8 = 24 + 8 = 32 \] 2. Calculate \(F(0)\): \[

To evaluate the definite integral of the function (3x + 2) from 0 to 4, we can follow the process of finding the antiderivative of the function first.

The function (3x + 2) can be integrated using the power rule. The antiderivative of (3x) is (\frac{3}{2}x^2), and the antiderivative of the constant (2) is (2x). Therefore, the combined antiderivative of the function (3x + 2) is:

[

F(x) = \frac{3}{2}x^2 + 2x

]

Next, we will evaluate (F(x)) at the bounds (0) and (4), and then subtract the value at the lower limit from the value at the upper limit:

1. Calculate (F(4)):

[

F(4) = \frac{3}{2}(4^2) + 2(4) = \frac{3}{2}(16) + 8 = 24 + 8 = 32

]

2. Calculate (F(0)):

[