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Math  /  Calculus

Question1. ∫25(−3v+4)dv\int_{2}^{5}(-3 v+4) d v
2. ∫−11(t2−2)dt\int_{-1}^{1}\left(t^{2}-2\right) d t
3. ∫−11(t3−9t)dt\int_{-1}^{1}\left(t^{3}-9 t\right) d t
4. ∫12(3x2−1)dx\int_{1}^{2}\left(\frac{3}{x^{2}}-1\right) d x
5. ∫−33v1/3dx\int_{-3}^{3} v^{1 / 3} d x
6. ∫026xdx\int_{0}^{2} 6 x d x

Studdy Solution

STEP 1

1. We are dealing with definite integrals, which involve integrating a function between two specific limits.
2. The integrals can be solved using standard integration techniques and properties of definite integrals.
3. The definite integrals will yield real numbers as results.

STEP 2

1. Solve ∫25(−3v+4)dv\int_{2}^{5}(-3v+4) dv.
2. Solve ∫−11(t2−2)dt\int_{-1}^{1}(t^{2}-2) dt.
3. Solve ∫−11(t3−9t)dt\int_{-1}^{1}(t^{3}-9t) dt.
4. Solve ∫12(3x2−1)dx\int_{1}^{2}(\frac{3}{x^2}-1) dx.
5. Solve ∫−33v1/3dx\int_{-3}^{3} v^{1/3} dx.
6. Solve ∫026xdx\int_{0}^{2} 6x dx.

STEP 3

Integrate the function −3v+4-3v+4 with respect to vv.
∫(−3v+4)dv=−3v22+4v+C \int (-3v+4) dv = -\frac{3v^2}{2} + 4v + C

STEP 4

Evaluate the definite integral from 22 to 55.
[−3v22+4v]25=(−3⋅522+4⋅5)−(−3⋅222+4⋅2) \left[ -\frac{3v^2}{2} + 4v \right]_{2}^{5} = \left( -\frac{3 \cdot 5^2}{2} + 4 \cdot 5 \right) - \left( -\frac{3 \cdot 2^2}{2} + 4 \cdot 2 \right)

STEP 5

Simplify the expression.
(−752+20)−(−122+8)=(−752+20)−(−6+8)=(−752+20)−2=−752+18=−75−362=−392 \left( -\frac{75}{2} + 20 \right) - \left( -\frac{12}{2} + 8 \right) = \left( -\frac{75}{2} + 20 \right) - \left( -6 + 8 \right) = \left( -\frac{75}{2} + 20 \right) - 2 = -\frac{75}{2} + 18 = -\frac{75 - 36}{2} = -\frac{39}{2}

STEP 6

Integrate the function t2−2t^2 - 2 with respect to tt.
∫(t2−2)dt=t33−2t+C \int (t^2 - 2) dt = \frac{t^3}{3} - 2t + C

STEP 7

Evaluate the definite integral from −1-1 to 11.
[t33−2t]−11=(133−2⋅1)−((−1)33−2⋅(−1)) \left[ \frac{t^3}{3} - 2t \right]_{-1}^{1} = \left( \frac{1^3}{3} - 2 \cdot 1 \right) - \left( \frac{(-1)^3}{3} - 2 \cdot (-1) \right)

STEP 8

Simplify the expression.
(13−2)−(−13+2)=(13−2)−(−13+2)=(−53)−(53)=−103 \left( \frac{1}{3} - 2 \right) - \left( \frac{-1}{3} + 2 \right) = \left( \frac{1}{3} - 2 \right) - \left( -\frac{1}{3} + 2 \right) = \left( -\frac{5}{3} \right) - \left( \frac{5}{3} \right) = -\frac{10}{3}

STEP 9

Integrate the function t3−9tt^3 - 9t with respect to tt.
∫(t3−9t)dt=t44−9t22+C \int (t^3 - 9t) dt = \frac{t^4}{4} - \frac{9t^2}{2} + C

STEP 10

Evaluate the definite integral from −1-1 to 11.
[t44−9t22]−11=(144−9⋅122)−((−1)44−9⋅(−1)22) \left[ \frac{t^4}{4} - \frac{9t^2}{2} \right]_{-1}^{1} = \left( \frac{1^4}{4} - \frac{9 \cdot 1^2}{2} \right) - \left( \frac{(-1)^4}{4} - \frac{9 \cdot (-1)^2}{2} \right)

STEP 11

Simplify the expression.
(14−92)−(14−92)=(14−184)−(14−184)=0 \left( \frac{1}{4} - \frac{9}{2} \right) - \left( \frac{1}{4} - \frac{9}{2} \right) = \left( \frac{1}{4} - \frac{18}{4} \right) - \left( \frac{1}{4} - \frac{18}{4} \right) = 0

STEP 12

Integrate the function 3x2−1\frac{3}{x^2} - 1 with respect to xx.
∫(3x2−1)dx=∫3x−2dx−∫1dx=−3x−1−x+C \int \left( \frac{3}{x^2} - 1 \right) dx = \int 3x^{-2} dx - \int 1 dx = -3x^{-1} - x + C

STEP 13

Evaluate the definite integral from 11 to 22.
[−3x−x]12=(−32−2)−(−3−1) \left[ -\frac{3}{x} - x \right]_{1}^{2} = \left( -\frac{3}{2} - 2 \right) - \left( -3 - 1 \right)

STEP 14

Simplify the expression.
(−32−2)−(−3−1)=(−32−2)+4=−32+2=12 \left( -\frac{3}{2} - 2 \right) - \left( -3 - 1 \right) = \left( -\frac{3}{2} - 2 \right) + 4 = -\frac{3}{2} + 2 = \frac{1}{2}

STEP 15

Integrate the function v1/3v^{1/3} with respect to xx.
∫v1/3dx \int v^{1/3} dx This step cannot be completed because the integrand does not involve the variable of integration, xx.

STEP 16

Integrate the function 6x6x with respect to xx.
∫6xdx=3x2+C \int 6x dx = 3x^2 + C

STEP 17

Evaluate the definite integral from 00 to 22.
[3x2]02=3⋅22−3⋅02=3⋅4=12 \left[ 3x^2 \right]_{0}^{2} = 3 \cdot 2^2 - 3 \cdot 0^2 = 3 \cdot 4 = 12
Solution Summary:
1. ∫25(−3v+4)dv=−392\int_{2}^{5}(-3v+4) dv = -\frac{39}{2}
2. ∫−11(t2−2)dt=−103\int_{-1}^{1}(t^2-2) dt = -\frac{10}{3}
3. ∫−11(t3−9t)dt=0\int_{-1}^{1}(t^3-9t) dt = 0
4. ∫12(3x2−1)dx=12\int_{1}^{2}(\frac{3}{x^2}-1) dx = \frac{1}{2}
5. ∫−33v1/3dx\int_{-3}^{3} v^{1/3} dx could not be evaluated due to a variable mismatch.
6. ∫026xdx=12\int_{0}^{2} 6x dx = 12

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