PLEASE HELP FIND DEFINITE INTEGRAL. I've been on this problem for an hour and a half and done it over 7 different ways. It can't be THIS hard!!!
∫ ((t^2)-t)^2) dt with b=-3 and a=1
thank you.
To find the definite integral of the function ((t^2)-t)^2 with respect to t, from a = 1 to b = -3, you can follow these steps:
Step 1: Expand the expression inside the squared term.
((t^2)-t)^2 = (t^2 - t)(t^2 - t)
Step 2: Multiply the terms using the distributive property.
(t^2 - t)(t^2 - t) = t^4 - t^3 - t^3 + t^2
Simplifying further yields:
t^4 - 2t^3 + t^2
Step 3: Now, find the indefinite integral of the simplified expression.
∫ (t^4 - 2t^3 + t^2) dt = (1/5)t^5 - (2/4)t^4 + (1/3)t^3 + C
Where C is the constant of integration.
Step 4: Evaluate the antiderivative at the upper limit (b = -3) and lower limit (a = 1) of integration.
Using the fundamental theorem of calculus, we subtract the antiderivative evaluated at a from the antiderivative evaluated at b.
[(1/5)(-3)^5 - (2/4)(-3)^4 + (1/3)(-3)^3] - [(1/5)(1)^5 - (2/4)(1)^4 + (1/3)(1)^3]
Step 5: Calculate the result.
[(1/5)(-243) - (2/4)(81) + (1/3)(-27)] - [(1/5)(1) - (2/4)(1) + (1/3)(1)]
= [-48.6 + 40.5 - 9] - [0.2 - 0.5 + 0.3]
= -48.6 + 40.5 - 9 - 0.2 + 0.5 - 0.3
= -17.1
Hence, the value of the definite integral ∫ ((t^2)-t)^2 dt from b = -3 to a = 1 is -17.1.