Math, asked by khanyousafzai830, 10 months ago

integration 3x^2(x^3-4)dx​

Answers

Answered by Anonymous
32

Question :

To evaluate

\sf\int3x^2(x^3-4)dx

Solution :

We have ,

\sf\int3x^2(x^3-4)dx

\sf\int\:(3x^5-12x^2)dx

\sf\int3x^5dx-\int12x^2dx

We know that

\sf\int\:x^n=\dfrac{x{}^{n+1}}{n+1}

Then,

\sf\int3x^5dx-\int12x^2dx

\sf=\dfrac{3x{}^{5+1}}{5+1}-\dfrac{12x{}^{2+1}}{2+1}+c

\sf=\dfrac{3x^6}{6}-\dfrac{12x^3}{3}+c

\sf=\dfrac{x^6}{2}-4x^3+c

\sf=\dfrac{x^6-8x^3}{2}+c

\rule{200}2

More About the topic :

The process of differentiation and integration are reverse of each other.

Properties of the indefinite integral

1) The indefinite integral of sum or difference of two function is equal to sum or difference of their integrals respectively.

\sf\int[f_1(x)\pm\:f_2(x)]dx=\int\:f_1(x)dx\pm\int\:f_2(x)dx

2) A constant factor of integrant may be taken outside the integral sign, if k is constant ,then

\sf\int\:kf(x)dx=k\int\:f(x)dx

Answered by Anonymous
0

Answer:

Question :

To evaluate

\sf\int3x^2(x^3-4)dx∫3x

2

(x

3

−4)dx

Solution :

We have ,

\sf\int3x^2(x^3-4)dx∫3x

2

(x

3

−4)dx

\sf\int\:(3x^5-12x^2)dx∫(3x

5

−12x

2

)dx

\sf\int3x^5dx-\int12x^2dx∫3x

5

dx−∫12x

2

dx

We know that

\sf\int\:x^n=\dfrac{x{}^{n+1}}{n+1}∫x

n

=

n+1

x

n+1

Then,

\sf\int3x^5dx-\int12x^2dx∫3x

5

dx−∫12x

2

dx

\sf=\dfrac{3x{}^{5+1}}{5+1}-\dfrac{12x{}^{2+1}}{2+1}+c=

5+1

3x

5+1

2+1

12x

2+1

+c

\sf=\dfrac{3x^6}{6}-\dfrac{12x^3}{3}+c=

6

3x

6

3

12x

3

+c

\sf=\dfrac{x^6}{2}-4x^3+c=

2

x

6

−4x

3

+c

\sf=\dfrac{x^6-8x^3}{2}+c=

2

x

6

−8x

3

+c

$$\rule{200}2$$

More About the topic :

The process of differentiation and integration are reverse of each other.

•Properties of the indefinite integral

1) The indefinite integral of sum or difference of two function is equal to sum or difference of their integrals respectively.

$$\sf\int[f_1(x)\pm\:f_2(x)]dx=\int\:f_1(x)dx\pm\int\:f_2(x)dx$$

2) A constant factor of integrant may be taken outside the integral sign, if k is constant ,then

$$\sf\int\:kf(x)dx=k\int\:f(x)dx$$

Step-by-step explanation:

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