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

Prepare for the JEE Main Integration Test with interactive quizzes and detailed explanations. Boost your integration skills, understand complex problems, and ace your exam. Master the dynamics of integration and put your knowledge to the test!

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

]

  1. Calculate (F(0)):

[

Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy