# How can I use integration by parts to find #int_{0}^{5}te^{-t}dt#?

You can do this:

By signing up, you agree to our Terms of Service and Privacy Policy

To find ( \int_{0}^{5} t \cdot e^{-t} , dt ) using integration by parts, we follow these steps:

- Let ( u = t ) and ( dv = e^{-t} , dt ).
- Compute ( du = dt ) and ( v = -e^{-t} ).
- Apply the integration by parts formula: ( \int u , dv = uv - \int v , du ).
- Plug in the values of ( u ), ( dv ), ( du ), and ( v ).
- Evaluate the definite integral from 0 to 5.

[ \int_{0}^{5} t \cdot e^{-t} , dt = \left[-t \cdot e^{-t}\right]*{0}^{5} - \int*{0}^{5} (-e^{-t}) , dt ]

[ = -5 \cdot e^{-5} - \left(-0 \cdot e^{-0}\right) - \left[-e^{-t}\right]_{0}^{5} ]

[ = -5 \cdot e^{-5} - 0 - \left(-e^{-5} + e^{0}\right) ]

[ = -5 \cdot e^{-5} + e^{-5} + 1 ]

[ = (1 - 5) \cdot e^{-5} + 1 ]

[ = -4 \cdot e^{-5} + 1 ]

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7