## Calculus: Early Transcendentals 8th Edition

$$h'(x) = xe^x$$
$$h(x) = \int\limits^{e^x}_1 {ln(t)dt}$$ Using FTC 1, substitue in the upper bound for t and multiply by the derivative of the upper bound. $$h'(x) = ln(e^x) \times (e^x)'$$ Simplify. [Note that $ln(e^x) = x$] $$h'(x) = xe^x$$