How do I solve these using U-Substitution?

Question

∫2 0 4xe^-x^2

∫4 0 2e^-x

∫-1 -4 -2e^x

∫2 1 2e^x

Answer 1

U substitution for exponential integrals usually follows a rule where you set u equal to the argument of the exponent.

1. a) ∫₀²4xe-^{x^2}dx
2. let u = -x²
3. du/dx = -2x
4. du =-2xdx treat the du/dx as a fraction and move dx to the other side.
5. -2du=4xdx this is almost the 4xdx from the original integral. Multiply by -2 to make 4xdx.
6. ∫₀² e^u(-2du) substitute line 4 into integral
7. -2∫₀² e^udu rewrite order for clarity
8. -2e^u|₀² use ∫ e^u du = e^u + C
9. -2e^{-x^2}|₀² substitute x back in place of u substitution
10. -2(e^-4 - 1) evaluate at limits. Use e^-0 = 1

b) use u = -x, -2du=2dx

1. ∫₀²2e^-xdx
2. let u = -x
3. du/dx = -1
4. du = -dx treat the du/dx as a fraction and move dx to the other side.
5. -2du =2dx this is almost the 2dx from the original integral. Multiply by -2 to make 2dx.
6. ∫₀² e^u(-2du) substitute line 4 into integral
7. -2∫₀² e^udu rewrite order for clarity
8. -2e^u|₀² use ∫ e^u du = e^u + C
9. -2e^{-x^2}|₀² substute x back in place of u substitution
10. -2(e^-4 - 1) evaluate at limits. Use e^-0 = 1

c)

1. ∫_-4^-1 -2e^xdx don’t need u substation. Use ∫ e^u du = e^u + C.

d)

1. ∫₁² 2e^xdx ∫ e^u du = e^u + C.

Answer 2

If I am reading correctly, I believe the first two are the only ones who would really benefit from u substitution. Any time I integrate e raised to the "something" power, I will assign that something to be my u. In the first case, u would be equal to -x^2. The last two appear to just contain e^x and so I would just leave it like that and integrate.