Evaluate the definite integral using the properties of even and odd functions.
S 2 (1/2 t^4+3)dt
-2
(1/10) t^5 + 3 t
both are odd so the answer from -2 to +2 is twice answer from 0 to 2
2 [ 2^5/10 + 6 ]
2 ( 9.2) = 18.4
Evaluate the definite integral using the properties of even and odd functions.
S 2 (1/2 t^4+3)dt
-2
I wonder why we call this PRE-calculus :)
To evaluate the definite integral ∫[-2, 2] (1/2 t^4 + 3) dt using the properties of even and odd functions, we can take advantage of the fact that an even function is symmetric about the y-axis, while an odd function is symmetric about the origin.
Step 1: Rewrite the integral using even and odd functions.
Split the integral into two parts: one involving the even function and the other involving the odd function.
∫[-2, 2] (1/2 t^4 + 3) dt
= ∫[-2, 2] 1/2 t^4 dt + ∫[-2, 2] 3 dt
Step 2: Evaluate the integral of the even function.
Since t^4 is an even function, we can simplify the first integral:
∫[-2, 2] 1/2 t^4 dt = 1/2 ∫[-2, 2] t^4 dt
Step 3: Use the symmetry property of even functions.
Because t^4 is an even function, we know that its integral over a symmetric interval (such as [-2, 2]) will be equal when integrated from -a to a, where "a" is the upper limit of integration.
∫[-2, 2] 1/2 t^4 dt = 1/2 ∫[0, 2] t^4 dt
= 1/2 [∫[0, 2] t^4 dt + ∫[-2, 0] t^4 dt]
= 1/2 [∫[0, 2] t^4 dt - ∫[0, 2] t^4 dt]
= 0
Step 4: Evaluate the integral of the odd function.
Next, we evaluate the second integral:
∫[-2, 2] 3 dt = 3 ∫[-2, 2] dt
Step 5: Use the symmetry property of odd functions.
Since the function is odd, it is symmetric about the origin. This means that the integral from -a to a will be equal to zero.
∫[-2, 2] 3 dt = 3 ∫[-2, 2] dt = 3 * 0 = 0
Step 6: Combine the results.
Finally, add the results from the two integrals:
∫[-2, 2] (1/2 t^4 + 3) dt = 0 + 0 = 0
Therefore, the definite integral ∫[-2, 2] (1/2 t^4 + 3) dt evaluated using the properties of even and odd functions is equal to 0.