3.7 Improper integrals  (Page 5/6)

${\displaystyle {\int }_2^4\frac{dx}{{(x-3)}^2}}$

${\displaystyle {\int }_0^{\infty }\frac{1}{4+x^2}}dx$

${\displaystyle {\int }_0^2\frac{1}{\sqrt{4-x^2}}dx}$

${\displaystyle {\int }_1^{\infty }\frac{1}{x\text{ln}x}dx}$

${\displaystyle {\int }_1^{\infty }x{e}^{\text{−}x}dx}$

${\displaystyle {\int }_{\text{−}\infty }^{\infty }\frac{x}{x^2+1}dx}$

Without integrating, determine whether the integral ${\displaystyle {\int }_1^{\infty }\frac{1}{\sqrt{x^3+1}}dx}$ converges or diverges by comparing the function $f(x)=\frac{1}{\sqrt{x^3+1}}$ with $g(x)=\frac{1}{\sqrt{x^3}}.$

Without integrating, determine whether the integral ${\displaystyle {\int }_1^{\infty }\frac{1}{\sqrt{x+1}}dx}$ converges or diverges.

${\displaystyle {\int }_0^{\infty }{e}^{\text{−}x}\text{cos}xdx}$

${\displaystyle {\int }_1^{\infty }\frac{\text{ln}x}{x}dx}$

${\displaystyle {\int }_0^1\frac{\text{ln}x}{\sqrt{x}}dx}$

${\displaystyle {\int }_0^1\text{ln}xdx}$

${\displaystyle {\int }_{\text{−}\infty }^{\infty }\frac{1}{x^2+1}dx}$

${\displaystyle {\int }_1^5\frac{dx}{\sqrt{x-1}}}$

${\displaystyle {\int }_{\mathrm{-2}}^2\frac{dx}{{\left(1+x\right)}^2}}$

${\displaystyle {\int }_0^{\infty }{e}^{\text{−}x}dx}$

${\displaystyle {\int }_0^{\infty }\text{sin}xdx}$

${\displaystyle {\int }_{\text{−}\infty }^{\infty }\frac{e^x}{1+e^{2x}}dx}$

${\displaystyle {\int }_0^1\frac{dx}{\sqrt[3]{x}}}$

${\displaystyle {\int }_0^2\frac{dx}{x^3}}$

${\displaystyle {\int }_{\mathrm{-1}}^2\frac{dx}{x^3}}$

${\displaystyle {\int }_0^1\frac{dx}{\sqrt{1-x^2}}}$

${\displaystyle {\int }_0^3\frac{1}{x-1}dx}$

${\displaystyle {\int }_1^{\infty }\frac{5}{x^3}dx}$

${\displaystyle {\int }_3^5\frac{5}{{(x-4)}^2}dx}$

${\displaystyle {\int }_1^{\infty }\frac{dx}{x^2+4x}};$ compare with ${\displaystyle {\int }_1^{\infty }\frac{dx}{x^2}}.$

${\displaystyle {\int }_1^{\infty }\frac{dx}{\sqrt{x}+1}};$ compare with ${\displaystyle {\int }_1^{\infty }\frac{dx}{2\sqrt{x}}}.$

${\displaystyle {\int }_1^{\infty }\frac{dx}{x^e}}$

${\displaystyle {\int }_0^1\frac{dx}{x^{\pi }}}$

${\displaystyle {\int }_0^1\frac{dx}{\sqrt{1-x}}}$

${\displaystyle {\int }_0^1\frac{dx}{1-x}}$

${\displaystyle {\int }_{\text{−}\infty }^0\frac{dx}{x^2+1}}$

${\displaystyle {\int }_{\mathrm{-1}}^1\frac{dx}{\sqrt{1-x^2}}}$

${\displaystyle {\int }_0^1\frac{\text{ln}x}{x}dx}$

${\displaystyle {\int }_0^e\text{ln}(x)dx}$

${\displaystyle {\int }_0^{\infty }x{e}^{\text{−}x}dx}$

${\displaystyle {\int }_{\text{−}\infty }^{\infty }\frac{x}{{\left(x^2+1\right)}^2}dx}$

${\displaystyle {\int }_0^{\infty }{e}^{\text{−}x}dx}$

${\displaystyle {\int }_0^9\frac{dx}{\sqrt{9-x}}}$

${\displaystyle {\int }_{\mathrm{-27}}^1\frac{dx}{x^{2\text{/}3}}}$

${\displaystyle {\int }_0^3\frac{dx}{\sqrt{9-x^2}}}$

${\displaystyle {\int }_6^{24}\frac{dt}{t\sqrt{t^2-36}}}$

${\displaystyle {\int }_0^{4}x\text{ln}(4x)dx}$

${\displaystyle {\int }_0^3\frac{x}{\sqrt{9-x^2}}dx}$

Evaluate ${\displaystyle {\int }_{.5}^t\frac{dx}{\sqrt{1-x^2}}}.$ (Be careful!) (Express your answer using three decimal places.)

Evaluate ${\displaystyle {\int }_1^4\frac{dx}{\sqrt{x^2-1}}}.$ (Express the answer in exact form.)

Evaluate ${\displaystyle {\int }_2^{\infty }\frac{dx}{{(x^2-1)}^{3\text{/}2}}}.$

Find the area of the region in the first quadrant between the curve $y=e^{\mathrm{-6}x}$ and the x -axis.

Find the area of the region bounded by the curve $y=\frac{7}{x^2},$ the x -axis, and on the left by $x=1.$

Find the area under the curve $y=\frac{1}{{\left(x+1\right)}^{3\text{/}2}},$ bounded on the left by $x=3.$

Find the area under $y=\frac{5}{1+x^2}$ in the first quadrant.

Find the volume of the solid generated by revolving about the x -axis the region under the curve $y=\frac{3}{x}$ from $x=1$ to $x=\infty .$

Find the volume of the solid generated by revolving about the y -axis the region under the curve $y=6e^{\mathrm{-2}x}$ in the first quadrant.

Find the volume of the solid generated by revolving about the x -axis the area under the curve $y=3e^{\text{−}x}$ in the first quadrant.

$f(x)=1$

$f(x)=x$

$f(x)=\text{cos}(2x)$

$f(x)=e^{ax}$

Use the formula for arc length to show that the circumference of the circle $x^2+y^2=1$ is $2\pi .$

Show that $f(x)=\{\begin{array}{c}0\text{if}x<0\\ 7e^{\mathrm{-7}x}\text{if}x\ge 0\end{array}$ is a probability density function.

Find the probability that x is between 0 and 0.3. (Use the function defined in the preceding problem.) Use four-place decimal accuracy.

${\displaystyle \int e^x\text{sin}(x)dx}$ cannot be integrated by parts.

${\displaystyle \int \frac{1}{x^4+1}dx}$ cannot be integrated using partial fractions.

In numerical integration, increasing the number of points decreases the error.

Integration by parts can always yield the integral.

${\displaystyle \int x^2\text{sin}(4x)dx}$ using integration by parts

${\displaystyle \int \frac{1}{x^2\sqrt{x^2+16}}dx}$ using trigonometric substitution

${\displaystyle \int \sqrt{x}\text{ln}(x)dx}$ using integration by parts

${\displaystyle \int \frac{3x}{x^3+2x^2-5x-6}dx}$ using partial fractions

${\displaystyle \int \frac{x^5}{{\left(4x^2+4\right)}^{5\text{/}2}}dx}$ using trigonometric substitution

${\displaystyle \int \frac{\sqrt{4-{\text{sin}}^2(x)}}{{\text{sin}}^2(x)}\text{cos}(x)dx}$ using a table of integrals or a CAS

${\displaystyle \int {\text{sin}}^2(x){\text{cos}}^2(x)dx}$

${\displaystyle \int x^3\sqrt{x^2+2}dx}$

${\displaystyle \int \frac{3x^2+1}{x^4-2x^3-x^2+2x}dx}$

${\displaystyle \int \frac{1}{x^4+4}dx}$

${\displaystyle \int \frac{\sqrt{3+16x^4}}{x^4}dx}$

[T] ${\displaystyle {\int }_1^2\sqrt{x^5+2}}dx$

[T] ${\displaystyle {\int }_0^{\sqrt{\pi }}{e}^{\text{−}\text{sin}(x^2)}}dx$

[T] ${\displaystyle {\int }_1^4\frac{\text{ln}\left(1\text{/}x\right)}{x}}dx$

${\displaystyle {\int }_1^{\infty }\frac{1}{x^n}}dx,$ for what values of $n$ does this integral converge or diverge?

${\displaystyle {\int }_1^{\infty }\frac{e^{\text{−}x}}{x}}dx$

Show that $\text{Γ}(a)=(a-1)\text{Γ}(a-1).$

Extend to show that $\text{Γ}(a)=(a-1)\text{!},$ assuming $a$ is a positive integer.

The fastest car in the world, the Bugati Veyron, can reach a top speed of 408 km/h. The graph represents its velocity.

[T] Use the graph to estimate the velocity every 20 sec and fit to a graph of the form $v(t)=a{\text{exp}}^{bx}\text{sin}(cx)+d.$ ( Hint: Consider the time units.)

[T] Using your function from the previous problem, find exactly how far the Bugati Veyron traveled in the 1 min 40 sec included in the graph.