Solution Found!
Solved: ?Improper Integrals that Are Both Type 1 and Type 2 The integral
Chapter 7, Problem 67(choose chapter or problem)
Improper Integrals that Are Both Type 1 and Type 2 The integral \(\int_{a}^{\infty} f(x) d x\) is improper because the interval \([a, \infty)\). If f has an infinite discontinuity at a, then the integral is improper for a second reason. In this case we evaluate the integral by expressing it as a sum of improper integrals of Type 2 and Type 1 as follows:
\(\int_{a}^{\infty} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{\infty} f(x) d x \quad c>a\)
Evaluate the given integral if it is convergent.
\(\int_{0}^{\infty} \frac{1}{\sqrt{x}(1+x)} d x\)
Questions & Answers
QUESTION:
Improper Integrals that Are Both Type 1 and Type 2 The integral \(\int_{a}^{\infty} f(x) d x\) is improper because the interval \([a, \infty)\). If f has an infinite discontinuity at a, then the integral is improper for a second reason. In this case we evaluate the integral by expressing it as a sum of improper integrals of Type 2 and Type 1 as follows:
\(\int_{a}^{\infty} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{\infty} f(x) d x \quad c>a\)
Evaluate the given integral if it is convergent.
\(\int_{0}^{\infty} \frac{1}{\sqrt{x}(1+x)} d x\)
ANSWER:Step 1 of 2
We can write
