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Could you check whether I am doing these questions right:

1. [itex] \int_{0}^{4} (2+x) dx [/itex]. So I use the Fundamental Theorem of Calculus [itex] F(b)-F(a)[/itex] and receive: [itex] \frac{(x+2)^{2}}{2} = F(4) - F(0) = 16 [/itex]

2. [itex] \int_{-1}^{1} (4t^{3} - 2t) dt = t^{4} - t^{2} = F(b)-F(a) = 0 [/itex]

3. [itex] \int_{0}^{3} \frac{1}{\sqrt{1+x}} dx = \frac{-1}{2}(1+x)^\frac{-3}{2} = F(b) - F(a) = \frac{7}{16} [/itex]

4. [itex] \int_{1}^{2}(\frac{1}{x^{2}} - \frac{1}{x^{3}}dx = \frac{x^{-1}}{-1} - \frac{x^{-2}}{-2} = F(b) - F(a) = \frac{1}{8}[/itex]

5. How would you do this one: [itex] \frac{3+ \ln x}{x} dx [/itex]?

6. [itex] \int^{1}_{-1} 3xe^{x^{2} -1} dx [/itex] Also how would I set this up? Would I let [itex] u = x^{2} - 1 [/itex]?

If the marginal cost is [itex] \frac{dC}{dx} = 675 + 0.5x [/itex] how would C change when [itex] x [/itex] increases from 50 to 51? So [itex] C = 675x + \frac{1}{4} x^{2} [/itex]. So would I just compute [itex] F(51) - F(50) [/itex]?

If you want to find the average value of [itex] f(x) = \frac{4}{\sqrt{x-1}}, [5,10] [/itex] would you use the formula [itex] \frac{1}{b-a}f'(x) [/itex]?

Thanks

1. [itex] \int_{0}^{4} (2+x) dx [/itex]. So I use the Fundamental Theorem of Calculus [itex] F(b)-F(a)[/itex] and receive: [itex] \frac{(x+2)^{2}}{2} = F(4) - F(0) = 16 [/itex]

2. [itex] \int_{-1}^{1} (4t^{3} - 2t) dt = t^{4} - t^{2} = F(b)-F(a) = 0 [/itex]

3. [itex] \int_{0}^{3} \frac{1}{\sqrt{1+x}} dx = \frac{-1}{2}(1+x)^\frac{-3}{2} = F(b) - F(a) = \frac{7}{16} [/itex]

4. [itex] \int_{1}^{2}(\frac{1}{x^{2}} - \frac{1}{x^{3}}dx = \frac{x^{-1}}{-1} - \frac{x^{-2}}{-2} = F(b) - F(a) = \frac{1}{8}[/itex]

5. How would you do this one: [itex] \frac{3+ \ln x}{x} dx [/itex]?

6. [itex] \int^{1}_{-1} 3xe^{x^{2} -1} dx [/itex] Also how would I set this up? Would I let [itex] u = x^{2} - 1 [/itex]?

If the marginal cost is [itex] \frac{dC}{dx} = 675 + 0.5x [/itex] how would C change when [itex] x [/itex] increases from 50 to 51? So [itex] C = 675x + \frac{1}{4} x^{2} [/itex]. So would I just compute [itex] F(51) - F(50) [/itex]?

If you want to find the average value of [itex] f(x) = \frac{4}{\sqrt{x-1}}, [5,10] [/itex] would you use the formula [itex] \frac{1}{b-a}f'(x) [/itex]?

Thanks

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