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Chapter 6: Problem 18

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{0}^{\infty}(x-1) e^{-x} d x $$

Short Answer

Expert verified
The given improper integral converges and its value is 1.

Step by step solution

01

Integration by Parts

Start by applying the integration by parts formula which is \(\int udv = uv - \int vdu\). First, set \(u = x-1\) and \(dv = e^{-x} dx\). Then we get, \(du = dx\) and \(v = -e^{-x}\). Apply the formula to get \[-e^{-x}(x-1) - \int -e^{-x} dx.\]
02

Evaluation

Evaluate the new integral, we get \[-e^{-x}(x-1) + e^{-x}.\] This is the integral from 0 to infinity.
03

Apply Limits

Now apply the limits, we substitute the upper and then the lower limits to get \[\lim_{t \to \infty} [-e^{-t}(t-1) + e^{-t}] - [-e^{0}(0-1) + e^{0}].\]
04

Simplification

Simplify the above expression, we get \[\lim_{t \to \infty} [-e^{-t}(t-1) + e^{-t}] + 1.\]
05

Limit at Infinity

Now, take the limit as \(t \to \infty\). Using L'Hopital's rule or knowledge about exponential functions, we know that \(-e^{-t}t\) and \(e^{-t}\) both approach zero as \(t \to \infty\). Therefore, the limit at infinity is \[0 + 1 = 1.\] So, this improper integral converges and its value is 1.

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Most popular questions from this chapter

The Gamma Function \(\Gamma(n)\) is defined by \(\Gamma(n)=\int_{0}^{\infty} x^{n-1} e^{-x} d x, \quad n>0\) (a) Find \(\Gamma(1), \Gamma(2),\) and \(\Gamma(3)\). (b) Use integration by parts to show that \(\Gamma(n+1)=n \Gamma(n)\). (c) Write \(\Gamma(n)\) using factorial notation where \(n\) is a positive integer.

(a) Let \(f^{\prime}(x)\) be continuous. Show that \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2 h}=f^{\prime}(x)\) (b) Explain the result of part (a) graphically.

Find the integral. Use a computer algebra system to confirm your result. $$ \int \frac{1-\sec t}{\cos t-1} d t $$

Use a computer algebra system to evaluate the definite integral. $$ \int_{0}^{\pi / 4} \sin 2 \theta \sin 3 \theta d \theta $$

Let \(f^{\prime \prime}(x)\) be continuous. Show that $$ \lim _{h \rightarrow 0} \frac{f(x+h)-2 f(x)+f(x-h)}{h^{2}}=f^{\prime \prime}(x) $$
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