Introduction to solving calculus math problem:
There are various branches exist in mathematics. The calculus is an important part of mathematics. It deals with the limits, functions, derivatives, integrals, and infinite series. In general calculus is divided into two parts,they are differential and integral calculus. Now we are going to discuss various problems related to calculus.
Example Problems for Solving Calculus Math Problem:
Solving calculus math problem – Example: 1
Solve `\int x^2e^xdx`
Solution:
Both side integration by the parts of twice to get this to an integral for which we know the formula.
`u=x^2`
`du=2xdx`
`dv=e^xdx`
`v=e^x`
`\int x^2e^xdx=x^2e^x-2\int xe^xdx`
`u=x`
`du=dx`
`dv=e^xdx`
`v=e^x`
`\int x^2e^x=x^2e^x-2\[xe^x-\int e^xdx\]=x^2e^x-2xe^x+2e^x+C`
Solving calculus math problem – Example: 2
Solve `\int \frac{\sqrt{x^2-4}}{x}dx`
Solution:
`x=2\sec \theta`
`dx=2\sec \theta\tan \thetad\theta`
`\int \frac{\sqrt{x^2-1}}{x}dx=\int \frac{\sqrt{4\sec^2\theta-4}}{2\sec \theta}(2)\sec \theta\tan \thetad\theta=2\int \sqrt{\tan^2\theta}\tan \thetad\theta=2\int \tan^2\thetad\theta`
We must use a trig substitution to get this in a form that is more easily integrable.
`2\int \tan^2\thetad\theta=2\int (\sec^2\theta-1)d\theta=2[\tan \theta-\theta]+C`
Now, we know `\sec \theta=\frac{x}{2} so \tan \theta=\frac{\sqrt{x^2-4}}{2} and \theta=\arcsec \frac{x}{2}.` So
`\int \frac{\sqrt{x^2-4}}{x}dx=2\ [\frac{\sqrt{x^2-4}}{2}-\arcsec \frac{x}{2}\ ]+C=\sqrt{x^2-4}-2\arcsec \frac{x}{2}+C`
Practice Problems for Solving Calculus Math Problem:
1. Solve `\int_0^\infty x^4 e^{-x^3} dx`
`Answer: \frac{2}{9}\Gamma\(\frac{2}{3}\)`
2. Simplify `\int_0^\infty 3^{-4z^2}dz`
`Answer: \sqrt{\frac{\pi}{16\ln3}}`
There are various branches exist in mathematics. The calculus is an important part of mathematics. It deals with the limits, functions, derivatives, integrals, and infinite series. In general calculus is divided into two parts,they are differential and integral calculus. Now we are going to discuss various problems related to calculus.
Example Problems for Solving Calculus Math Problem:
Solving calculus math problem – Example: 1
Solve `\int x^2e^xdx`
Solution:
Both side integration by the parts of twice to get this to an integral for which we know the formula.
`u=x^2`
`du=2xdx`
`dv=e^xdx`
`v=e^x`
`\int x^2e^xdx=x^2e^x-2\int xe^xdx`
`u=x`
`du=dx`
`dv=e^xdx`
`v=e^x`
`\int x^2e^x=x^2e^x-2\[xe^x-\int e^xdx\]=x^2e^x-2xe^x+2e^x+C`
Solving calculus math problem – Example: 2
Solve `\int \frac{\sqrt{x^2-4}}{x}dx`
Solution:
`x=2\sec \theta`
`dx=2\sec \theta\tan \thetad\theta`
`\int \frac{\sqrt{x^2-1}}{x}dx=\int \frac{\sqrt{4\sec^2\theta-4}}{2\sec \theta}(2)\sec \theta\tan \thetad\theta=2\int \sqrt{\tan^2\theta}\tan \thetad\theta=2\int \tan^2\thetad\theta`
We must use a trig substitution to get this in a form that is more easily integrable.
`2\int \tan^2\thetad\theta=2\int (\sec^2\theta-1)d\theta=2[\tan \theta-\theta]+C`
Now, we know `\sec \theta=\frac{x}{2} so \tan \theta=\frac{\sqrt{x^2-4}}{2} and \theta=\arcsec \frac{x}{2}.` So
`\int \frac{\sqrt{x^2-4}}{x}dx=2\ [\frac{\sqrt{x^2-4}}{2}-\arcsec \frac{x}{2}\ ]+C=\sqrt{x^2-4}-2\arcsec \frac{x}{2}+C`
Practice Problems for Solving Calculus Math Problem:
1. Solve `\int_0^\infty x^4 e^{-x^3} dx`
`Answer: \frac{2}{9}\Gamma\(\frac{2}{3}\)`
2. Simplify `\int_0^\infty 3^{-4z^2}dz`
`Answer: \sqrt{\frac{\pi}{16\ln3}}`
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