- #1

- 383

- 5

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Superposed_Cat
- Start date

- #1

- 383

- 5

- #2

- 3

- 0

Now reverse the differentiation.

So integrating [itex]3x^2[/itex] you first increase the power and then divide by the new power. So...

1) Raise the power: [itex]3x^3[/itex]

2) Divide by the new power: [itex]\frac{3}{3}x^3[/itex] = [itex]x^3[/itex]

That's all that's happening on the second line.

- #3

- 575

- 76

##\int 3x^\frac{3}{4}\ dx+\int 7x^{-5}\ dx+\int \frac{1}{6}x^{-\frac{1}{2}}\ dx=3\int x^\frac{3}{4}\ dx+7\int x^{-5}\ dx+\frac{1}{6}\int x^{-\frac{1}{2}}\ dx## using the constant multiple rule for integrals

##3\int x^\frac{3}{4}\ dx+7\int x^{-5}\ dx+\frac{1}{6}\int x^{-\frac{1}{2}}\ dx=3\cdot\frac{1}{\frac{3}{4}+1} x^{\frac{3}{4}+1}+7\cdot\frac{1}{-5+1}x^{-5+1}+\frac{1}{6}\cdot\frac{1}{-\frac{1}{2}+1} x^{-\frac{1}{2}+1}+c## using the integral formula ##\int x^\alpha\ dx=\frac{1}{a+1}x^{\alpha+1}+c## when ##\alpha\neq-1##

- #4

- 383

- 5

Profuse thanks dear sir.

Share:

- Replies
- 3

- Views
- 2K

- Replies
- 2

- Views
- 327