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Problem 2 Easy Difficulty
Which of the following integrals are improper? Why?
(a) $ \displaystyle \int_0^{\frac{\pi}{4}} \tan x\ dx $
(b) $ \displaystyle \int_0^\pi \tan x\ dx $
(c) $ \displaystyle \int_{-1}^1 \frac{dx}{x^2 - x - 2} $
(d) $ \displaystyle \int_0^\infty e^{-x^3}\ dx $
Answer
a. $[0, \pi / 4]$
b. $\lim _{a \rightarrow-\pi / 2} \int_{0}^{a} \tan x d x+\lim _{b \rightarrow+\pi / 2} \int_{b}^{\pi} \tan x d x$
c. $\lim _{a \rightarrow+(-1)} \int_{a}^{1} \frac{1}{x^{2}-x-2}$
d. $\lim _{a \rightarrow \infty} \int_{0}^{a} e^{-x^{3}} d x$
Topics
Integration Techniques
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so far, eh? Look at the graph ofthe ten and ice. First interval from zero to pie before is finite and there's a function of ten attacks has no infinite. It is continued on the interval from their tow pilot for so it is not improper. B also look at it A graph of ten an axe Wei will find Han and pile off too. Yes, infinite. So hand and eggs has infinite is continually on this angel from there to pie. So it is improper Integral. See what axe goes to make to Juan. Denominator of dysfunction. I like square minus X minus two goes to zero. This function got to infinite. What fax goes to get you one So it is improper for the this into a wall is given it so it is improper, integral.
Topics
Integration Techniques
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