Math  /  Calculus

QuestionDetermine the following indefinite integrals:
1. 25(3v+4)dv\int_{2}^{5}(-3 v+4) d v
2. 11(t22)dt\int_{-1}^{1}\left(t^{2}-2\right) d t
3. 11(t39t)dt\int_{-1}^{1}\left(t^{3}-9 t\right) d t
4. 12(3x21)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
Subtract the value at the lower limit from the value at the upper limit for the sixth integral.
120 12 - 0 =12 = 12
Solutions:
1. 25(3v+4)dv=19.5\int_{2}^{5}(-3v + 4) \, dv = -19.5
2. 11(t22)dt=103\int_{-1}^{1}(t^2 - 2) \, dt = -\frac{10}{3}
3. 11(t39t)dt=0\int_{-1}^{1}(t^3 - 9t) \, dt = 0
4. 12(3x21)dx=12\int_{1}^{2}\left(\frac{3}{x^2} - 1\right) \, dx = \frac{1}{2}
5. 33v1/3dv=0\int_{-3}^{3} v^{1/3} \, dv = 0
6. 026xdx=12\int_{0}^{2} 6x \, dx = 12

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