Integral 09.11.08

November 8, 2009 By វិចិត្រ 3 Comments

Find the following integrals

1.	$\int \left(6x^2+8x+3\right)dx$
2.	$\int x(x+a)(x+b)dx$

Solution

1. We have:
$\begin{array}{ll} \int \left(6x^2+8x+3\right)dx &= \int 6x^2 dx + \int 8x dx +\int 3dx \\ &=6\int x^2 dx +8 \int xdx +3 \int dx \\ &=6\left[\frac{1}{2+1}x^{2+1}+C_1\right] +8 \left[\frac{1}{1+1}x^{1+1}+C_2\right]+3\left[\frac{1}{0+1}x^{0+1}+C_3\right] \\ &= 2x^3+4x^2+3x+(C_1+C_2+C_3)\\ &=2x^3+4x^2+3x+C \end{array}$
where $C=C_1+C_2+C_3$ is a constant.

2. We have:
$\begin{array}{ll} \int x(x+a)(x+b)dx &= \int x\left(x^2 +(a+b)x+ab\right)dx \\ &= \int x^3+(a+b)x^2+abx dx \\ &=\frac{1}{3+1} x^{3+1}+ \frac{a+b}{2+1} x^{2+1}+\frac{ab}{1+1} x^{1+1} +C \\ &= \frac{1}{4} x^4+\frac{a+b}{3}x^3+\frac{ab}{2}x^2+C \end{array}$

Filed Under: គណិតវិទ្យា, អប់រំ Tagged With: integral

Comments

hay pahen says:

November 21, 2009 at 9:37 am

hello teacher.i am very happy when i show your website and your book.but now i want to ask you one qeustion that : solve eqution sin3x+2cosx-2=0.

thank you very much.

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lsvichet says:

November 21, 2009 at 11:30 am

Now i'm finding solution… I will inform you later

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lsvichet says:

December 7, 2009 at 4:03 pm

គណនាប៉ុន្មានថ្ងៃនៅតែរកចំលើយមិនទាន់ឃើញទៀត

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