Question

Evaluate the definite integral. Use a graphing utility to verify your result. $$ \int_{1}^{2} e^{1-x} d x $$

Short answer

The definite integral is \(1 - 1/e\).

Step-by-step solution

Identify the integral

Given the integral \(\int_{1}^{2} e^{1-x} dx\).

Integration

The integration of \(e^{1-x}\) results in \(-e^{1-x} + C\), where C is the constant of integration. This is due to the chain rule on the anti-derivative of the exponential function.

Application of the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if a function \(f\) is continuous over the interval [a, b] and \(F\) is an antiderivative of \(f\) on [a, b], then \(\int_{a}^{b} f(x) dx = F(b) - F(a)\). In this case, apply this theorem to obtain \((-(e^{1-2}) + C) - (-(e^{1-1}) + C)\).

Simplification

Simplify \((-(e^{1-2}) + C) - (-(e^{1-1}) + C)\) to get \(-e^{-1}-(-e^0)\). This simplifies further into \(1 - 1/e\).

Check with a graphing utility

Confirm this result using a graphing utility by graphing the integrand \(e^{1-x}\) and finding the area under the curve from 1 to 2.