Question

Let \(x>0\) and \(b>0 .\) Show that \(\int_{-b}^{b} e^{x t} d t=\frac{2 \sinh b x}{x}\).

Short answer

The integral \(\int_{-b}^{b} e^{x t} d t\) equals \(\frac{2 \sinh(bx)}{x}\)

Step-by-step solution

Compute the integral

The given problem involves the computation of a definite integral. The integral to be computed is \(\int_{-b}^{b} e^{x t} d t\). Start by substituting the upper and lower limits of integration into the exponential function and subtract these results from each other. This will yield \(e^{x b}-e^{-x b}\)

Simplify using the hyperbolic sine

The result, \(e^{x b}-e^{-x b}\), is a difference which can be realized as the definition of \(\sinh(x)\). The expression is double the hyperbolic sine of \(b x\): \(2\sinh(bx)\)

Divide the equation by 'x'

The equation from the previous step can be divided by \(x\), resulting in \(\frac{2 \sinh(bx)}{x}\). This is the solution to the problem.