Chapter 5: Problem 47
47-50. Find each integral. [Hint: Try some algebra.] $$ \int(x+1) x^{2} d x $$
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Chapter 5: Problem 47
47-50. Find each integral. [Hint: Try some algebra.] $$ \int(x+1) x^{2} d x $$
These are the key concepts you need to understand to accurately answer the question.
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Which one of these formulas is correct? a. \(\int \ln x d x=\frac{1}{|x|}+C\) b. \(\int \ln |x| d x=\frac{1}{x}+C\) c. \(\int \frac{1}{x} d x=\ln |x|+C\) d. \(\int \frac{1}{\ln x} d x=|x|+C\)
A friend says that if you can move numbers across the integral sign, you can do the same for variables since variables stand for numbers, and in this way you can always "fix" the differential \(d u\) to be what you want. Is your friend right?
In 2000 the birthrate in Africa increased from \(17 e^{0.02 t}\) million births per year to \(22 e^{0.02 t}\) million births per year, where \(t\) is the number of years since 2000 . Find the total increase in population that will result from this higher birthrate between \(2000(t=0)\) and \(2050(t=50)\).
\(85-94 .\) The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found as explained on the left (as well as by the methods of Section 6.1). $$ \int(x+1)(x-5)^{4} d x $$
Suppose that a company found its sales rate (in sales per day) if it did advertise, and also its (lower) sales rate if it did not advertise. If you integrated "upper minus lower" over a month, describe the meaning of the number that you would find.
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