Evaluate the definite integral.
function: (t+8)(t^2+3) with respect to variable t
lower limit: -sqrt(2)
upper limit: sqrt(2)
∫[-√2,√2] (t+8)(t^2+3) dt
= ∫[-√2,√2] t^3 + 8t^2 + 3t + 24 dt
= 1/4 t^4 + 8/3 t^3 + 3/2 t^2 + 24t [-√2,√2]
= 176/3 √2
Thank you!
To evaluate the definite integral of the function (t+8)(t^2+3) with respect to variable t over the interval from -sqrt(2) to sqrt(2), you can follow these steps:
1. First, expand the function using the distributive property: (t+8)(t^2+3) = t^3 + 3t + 8t^2 + 24.
2. Now, integrate each term of the expanded function separately. The integral of t^3 is (1/4)t^4, the integral of 3t is (3/2)t^2, the integral of 8t^2 is (8/3)t^3, and the integral of 24 is 24t.
3. Evaluate each term at the upper limit, sqrt(2), and subtract the result from the evaluation at the lower limit, -sqrt(2). Let's calculate each term separately:
- For the term (1/4)t^4, evaluate it at sqrt(2) and subtract the evaluation at -sqrt(2):
[(1/4)(sqrt(2))^4] - [(1/4)(-sqrt(2))^4] = (1/4)(2^2) - (1/4)(2^2) = 1 - 1 = 0.
- For the term (3/2)t^2, evaluate it at sqrt(2) and subtract the evaluation at -sqrt(2):
[(3/2)(sqrt(2))^2] - [(3/2)(-sqrt(2))^2] = (3/2)(2) - (3/2)(2) = 3 - 3 = 0.
- For the term (8/3)t^3, evaluate it at sqrt(2) and subtract the evaluation at -sqrt(2):
[(8/3)(sqrt(2))^3] - [(8/3)(-sqrt(2))^3] = (8/3)(2sqrt(2)) - (8/3)(-2sqrt(2)) = (16/3)(sqrt(2)) + (16/3)(sqrt(2)) = 32/3 sqrt(2).
- For the constant term 24t, evaluate it at sqrt(2) and subtract the evaluation at -sqrt(2):
24(sqrt(2)) - 24(-sqrt(2)) = 24(sqrt(2)) + 24(sqrt(2)) = 48 sqrt(2).
4. Finally, sum up all the evaluated terms:
0 + 0 + 32/3 sqrt(2) + 48 sqrt(2) = 32/3 sqrt(2) + 48 sqrt(2).
So, the value of the definite integral of the function (t+8)(t^2+3) with respect to t, from -sqrt(2) to sqrt(2), is 32/3 sqrt(2) + 48 sqrt(2).