Studdy Solution
STEP 1
1. We are tasked with finding the indefinite integrals of given functions.
2. Each integral will be evaluated over specified limits, making them definite integrals.
3. We will apply basic integration rules and the Fundamental Theorem of Calculus.
STEP 2
1. Evaluate the integral ∫25(−3v+4)dv.
2. Evaluate the integral ∫−11(t2−2)dt.
3. Evaluate the integral ∫−11(t3−9t)dt.
4. Evaluate the integral ∫12(x23−1)dx.
5. Evaluate the integral ∫−33v1/3dv.
6. Evaluate the integral ∫026xdx.
STEP 3
Evaluate ∫25(−3v+4)dv.
First, find the antiderivative:
∫(−3v+4)dv=−23v2+4v+C
STEP 4
Apply the limits of integration:
[−23(5)2+4(5)]−[−23(2)2+4(2)]
=[−23(25)+20]−[−23(4)+8]
=[−37.5+20]−[−6+8]
=(−17.5)−(2)
=−19.5
STEP 5
Evaluate ∫−11(t2−2)dt.
First, find the antiderivative:
∫(t2−2)dt=31t3−2t+C
STEP 6
Apply the limits of integration:
[31(1)3−2(1)]−[31(−1)3−2(−1)]
=[31−2]−[−31+2]
=[−35]−[35]
=−310
STEP 7
Evaluate ∫−11(t3−9t)dt.
First, find the antiderivative:
∫(t3−9t)dt=41t4−29t2+C
STEP 8
Apply the limits of integration:
[41(1)4−29(1)2]−[41(−1)4−29(−1)2]
=[41−29]−[41−29]
=0
STEP 9
Evaluate ∫12(x23−1)dx.
First, find the antiderivative:
∫(x23−1)dx=∫(3x−2−1)dx
=−3x−1−x+C
=−x3−x+C
STEP 10
Apply the limits of integration:
[−23−2]−[−13−1]
=[−23−2]−[−3−1]
=[−23−2]+4
=21
STEP 11
Evaluate ∫−33v1/3dv.
First, find the antiderivative:
∫v1/3dv=43v4/3+C
STEP 12
Apply the limits of integration:
Since v1/3 is an odd function and the limits are symmetric about zero, the integral evaluates to zero:
∫−33v1/3dv=0
STEP 13
Evaluate ∫026xdx.
First, find the antiderivative:
∫6xdx=3x2+C
STEP 14
Apply the limits of integration:
[3(2)2]−[3(0)2]
=12−0
=12
The solutions to the integrals are:
1. −19.5
2. −310
3. 0
4. 21
5. 0
6. 12