find the definite integral for 3x+pi for the fraction interval pi over 3 and pi over 6
interpretation:
find ∫(3x + π) dx from π/3 to π/6
= [(3/2)x^2 + πx] from π/3 to π/6
= (3/2)(π^2/36) + π(π/6) - ((3/2)(π^2/9) + π(π/3)
= π^2 /24 + π^2/6 - π^2 /6 - π^2/3
= -7π/24
How to find the definite integral for (3x+pi)cos3xdx for the fraction interval pi over 6 and pi over 3
http://www.wolframalpha.com/input/?i=%E2%88%AB(3x%2Bpi)(cos3x)+dx
To find the definite integral for the function `3x + pi` over the interval `[pi/3, pi/6]`, you can follow these steps:
Step 1: Find the antiderivative of the function.
To find the antiderivative, you need to integrate each term of the function with respect to `x` while treating constants, like `pi`, as coefficients:
∫ (3x + pi) dx = (3/2) x^2 + pi x + C,
where `C` is the constant of integration.
Step 2: Evaluate the definite integral.
To evaluate the definite integral, substitute the upper and lower limits of the integral into the antiderivative function and subtract the result evaluated at the lower limit from the result evaluated at the upper limit. In this case, the upper limit is `pi/6` and the lower limit is `pi/3`:
∫[pi/3, pi/6] (3x + pi) dx = [(3/2)(pi/6)^2 + pi(pi/6)] - [(3/2)(pi/3)^2 + pi(pi/3)].
Simplifying further,
= [(3/2)(pi^2/36) + (pi^2/6)] - [(3/2)(pi^2/9) + (pi^2/3)].
Now, we can simplify the expression.
= (pi^2/72 + pi^2/6) - (pi^2/6 + 2pi^2/9).
= (pi^2/72) - (2pi^2/9).
= pi^2/72 - 16pi^2/72.
= -15pi^2/72.
= -5pi^2/24.
Therefore, the definite integral of the function `3x + pi` over the interval `[pi/3, pi/6]` is `-5pi^2/24`.