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- https://math.libretexts.org/Courses/Cosumnes_River_College/Math_400%3A_Calculus_I_-_Differential_Calculus/00%3A_Front_Matter
- https://math.libretexts.org/Courses/Mission_College/Math_3B%3A_Calculus_2_(Sklar)/05%3A_Integration/5.07%3A_Integrals_Resulting_in_Inverse_Trigonometric_Functions/5.7E%3A_Exercises_for_Section_5.7Using the hint, one has \(\displaystyle ∫\frac{\csc^2x}{\csc^2x+\cot^2x}\,dx=∫\frac{\csc^2x}{1+2\cot^2x}\,dx.\) Set \(u=\sqrt{2}\cot x.\) Then, \(du=−\sqrt{2}\csc^2x\) and the integral is \(\displayst...Using the hint, one has \(\displaystyle ∫\frac{\csc^2x}{\csc^2x+\cot^2x}\,dx=∫\frac{\csc^2x}{1+2\cot^2x}\,dx.\) Set \(u=\sqrt{2}\cot x.\) Then, \(du=−\sqrt{2}\csc^2x\) and the integral is \(\displaystyle −\tfrac{1}{\sqrt{2}}∫\frac{du}{1+u^2}=−\tfrac{\sqrt{2}}{2}\tan^{−1}u+C=\tfrac{\sqrt{2}}{2}\tan^{−1}(\sqrt{2}\cot x)+C\).
- https://math.libretexts.org/Courses/Mission_College/Math_3B%3A_Calculus_2_(Sklar)/05%3A_Integration/5.05%3A_Substitution/5.5E%3A_Exercises_for_Section_5.5\(\displaystyle ∫^x_0g(t)\,dt=\frac{1}{2}∫^1_{u=1−x^2} \frac{du}{u^a}=\frac{1}{2(1−a)}u^{1−a}∣1u=\frac{1}{2(1−a)}(1−(1−x^2)^{1−a})\) As \(x→1\) the limit is \(\dfrac{1}{2(1−a)}\) if \(a<1\), and the l...\(\displaystyle ∫^x_0g(t)\,dt=\frac{1}{2}∫^1_{u=1−x^2} \frac{du}{u^a}=\frac{1}{2(1−a)}u^{1−a}∣1u=\frac{1}{2(1−a)}(1−(1−x^2)^{1−a})\) As \(x→1\) the limit is \(\dfrac{1}{2(1−a)}\) if \(a<1\), and the limit diverges to \(+∞\) if \(a>1\). 64) The area of the top half of an ellipse with a major axis that is the \(x\)-axis from \(x=−1\) to a and with a minor axis that is the \(y\)-axis from \(y=−b\) to \(y=b\) can be written as \(\displaystyle ∫^a_{−a}b\sqrt{1−\frac{x^2}{a^2}}\,dx\).
- https://math.libretexts.org/Courses/Mission_College/Math_3B%3A_Calculus_2_(Sklar)/05%3A_Integration/5.06%3A_Integrals_Involving_Exponential_and_Logarithmic_Functions/5.6E%3A_Exercises_for_Section_5.644) \(\displaystyle ∫\frac{\sin x+\cos x}{\sin x−\cos x}\,dx;\quad u=\sin x−\cos x\) \(\displaystyle ∫\frac{\sin x+\cos x}{\sin x−\cos x}\,dx \quad=\quad \ln|\sin x−\cos x|+C\) 60) Compute the integra...44) \(\displaystyle ∫\frac{\sin x+\cos x}{\sin x−\cos x}\,dx;\quad u=\sin x−\cos x\) \(\displaystyle ∫\frac{\sin x+\cos x}{\sin x−\cos x}\,dx \quad=\quad \ln|\sin x−\cos x|+C\) 60) Compute the integral of \(f(x)=xe^{−x^2}\) and find the smallest value of \(N\) such that the area under the graph \(f(x)=xe^{−x^2}\) between \( x=N\) and \(x=N+10\) is, at most, \(0.01\).
- https://math.libretexts.org/Courses/Cosumnes_River_College/Math_400%3A_Calculus_I_-_Differential_Calculus/05%3A_Investigating_Integrals/5.07%3A_Chapter_5_Review_Exercises24) The velocity of a bullet from a rifle can be approximated by \( v(t)=6400t^2−6505t+2686,\) where \( t\) is seconds after the shot and v is the velocity measured in feet per second. This equation o...24) The velocity of a bullet from a rifle can be approximated by \( v(t)=6400t^2−6505t+2686,\) where \( t\) is seconds after the shot and v is the velocity measured in feet per second. This equation only models the velocity for the first half-second after the shot: \( 0≤t≤0.5.\) What is the total distance the bullet travels in \(0.5\) sec?
- https://math.libretexts.org/Courses/City_College_of_San_Francisco/CCSF_Calculus/15%3A_Multiple_Integration/15.04%3A_Double_Integrals_in_Polar_Coordinates/15.4E%3A_Exercises_for_Section_15.350) The surface of a right circular cone with height \(h\) and base radius \(a\) can be described by the equation \(f(x,y)=h-h\sqrt{\frac{x^2}{a^2}+\frac{y^2}{a^2}}\), where the tip of the cone lies a...50) The surface of a right circular cone with height \(h\) and base radius \(a\) can be described by the equation \(f(x,y)=h-h\sqrt{\frac{x^2}{a^2}+\frac{y^2}{a^2}}\), where the tip of the cone lies at \((0,0,h)\) and the circular base lies in the \(xy\)-plane, centered at the origin.
- https://math.libretexts.org/Courses/City_College_of_San_Francisco/CCSF_Calculus/04%3A_Appropriate_Applications/4.06%3A_Optimization/4.6E%3A_Exercises1) When you find the maximum for an optimization problem, why do you need to check the sign of the derivative around the critical points? 16) You can run at a speed of \(6\) mph and swim at a speed of...1) When you find the maximum for an optimization problem, why do you need to check the sign of the derivative around the critical points? 16) You can run at a speed of \(6\) mph and swim at a speed of \(3\) mph and are located on the shore, \(4\) miles east of an island that is \(1\) mile north of the shoreline. The plush material for the square bottom of the box costs \($5/\text{ft}^2\) and the material for the sides costs \($2/\text{ft}^2\).
- https://math.libretexts.org/Courses/City_College_of_San_Francisco/CCSF_Calculus/06%3A_Applications_of_Integration/6.08%3A_Chapter_1_Review_Exercises1) The amount of work to pump the water out of a half-full cylinder is half the amount of work to pump the water out of the full cylinder. 20) The shape created by revolving the region between \(y=4+x...1) The amount of work to pump the water out of a half-full cylinder is half the amount of work to pump the water out of the full cylinder. 20) The shape created by revolving the region between \(y=4+x, \;y=3−x, \;x=0,\) and \(x=2\) rotated around the \(y\)-axis. 25) Find the volume of the catenoid \(y=\cosh(x)\) from \(x=−1\) to \(x=1\) that is created by rotating this curve around the \(x\)-axis, as shown here.
- https://math.libretexts.org/Courses/Cosumnes_River_College/Math_401%3A_Calculus_II_-_Integral_Calculus/01%3A_Applications_of_Integration/1.10%3A_Chapter_1_Review_Exercises1) The amount of work to pump the water out of a half-full cylinder is half the amount of work to pump the water out of the full cylinder. 20) The shape created by revolving the region between \(y=4+x...1) The amount of work to pump the water out of a half-full cylinder is half the amount of work to pump the water out of the full cylinder. 20) The shape created by revolving the region between \(y=4+x, \;y=3−x, \;x=0,\) and \(x=2\) rotated around the \(y\)-axis. 25) Find the volume of the catenoid \(y=\cosh(x)\) from \(x=−1\) to \(x=1\) that is created by rotating this curve around the \(x\)-axis, as shown here.
- https://math.libretexts.org/Courses/Cosumnes_River_College/Math_401%3A_Calculus_II_-_Integral_Calculus/02%3A_Techniques_of_Integration/2.03%3A_Integration_by_Parts/2.3E%3A_Exercises69) Find the volume of the solid generated by revolving the region bounded by the curve \(y=\ln x\), the \(x\)-axis, and the vertical line \(x=e^2\) about the \(x\)-axis. (Express the answer in exact ...69) Find the volume of the solid generated by revolving the region bounded by the curve \(y=\ln x\), the \(x\)-axis, and the vertical line \(x=e^2\) about the \(x\)-axis. (Express the answer in exact form.) 71) Find the volume of the solid generated by revolving the region in the first quadrant bounded by \(y=e^x\) and the \(x\)-axis, from \(x=0\) to \(x=\ln(7)\), about the \(y\)-axis. (Express the answer in exact form.)
- https://math.libretexts.org/Courses/Mission_College/Math_4A%3A_Multivariable_Calculus_v2_(Reed)/14%3A_Differentiation_of_Functions_of_Several_Variables/14.02%3A_Limits_and_Continuity/14.2E%3A_Exercises_for_Section_14.2\(\displaystyle \lim_{(x,y)→(a,b)}\left[\dfrac{2f(x,y) - 4g(x,y)}{f(x,y) - g(x,y)}\right] = \frac{2\left(\displaystyle \lim_{(x,y)→(a,b)}f(x,y)\right) - 4 \left(\displaystyle \lim_{(x,y)→(a,b)}g(x,y)\...\(\displaystyle \lim_{(x,y)→(a,b)}\left[\dfrac{2f(x,y) - 4g(x,y)}{f(x,y) - g(x,y)}\right] = \frac{2\left(\displaystyle \lim_{(x,y)→(a,b)}f(x,y)\right) - 4 \left(\displaystyle \lim_{(x,y)→(a,b)}g(x,y)\right)}{\displaystyle \lim_{(x,y)→(a,b)}f(x,y) - \displaystyle \lim_{(x,y)→(a,b)}g(x,y)}= \frac{2(5) - 4(2)}{5 - 2} = \frac{2}{3}\) 4) Show that the limit \(\displaystyle \lim_{(x,y)→(0,0)}\frac{5x^2y}{x^2+y^2}\) exists and is the same along the paths: \(y\)-axis and \(x\)-axis, and along \( y=x\).
