# How to find the integral of f(x) =x^3 - 5 between [ 1, 2]

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### 3 Answers

To find the integral of f(x) =x^3 - 5 between [ 1, 2], we first find the indefinite integral of f(x) = x^3 - 5

Int [ x^3 - 5]

=> Int [ x^3] - Int [5]

=> x^4 / 4 - 5x + C

Int [ f(x)] = x^4 / 4 - 5x + C

Now for x = 2, the value of Int [ f(x)] is 2^4 / 4 - 5*2 + C

For x = 1, the value of Int [ f(x)] is 1/4 - 5 + C

The required integral is 2^4 / 4 - 5*2 + C - (1/4 - 5 + C)

=> 4 - 10 + C - 1/4 + 5 - C

=> -5/4

**The required result is -5/4**

Given the curve f(x) = x^3 - 5

We need to find the definite integral on the interval [1, 2].

Let us intergrate f(x).

Let F(x) = intg f(x)

==> The defininte integral is :

I = F(2) - F(1).

==> F(x) = intg (x^3 - 5) dx

= intg x^3 dx - intg 5 dx

= x^4/4 - 5x + C

==> F(2) = 2^4/4 - 5*2 + C

= 4 - 10 + c = -6 + c

==> F(1) = 1/4 - 5 + C = -19/4 + C

==> I = -6 + 19/4 = ( -24 + 19) /4 = -5/4

**==> Then, the integral is I = -5/4**

To find the definite integral of f(x) = x^3 -5 for x= 1 to x= 2.

We know that If F(x) = Integral f(x) dx , then F(a) - F(a) = Int f(x) dx from x= a to x = b.

Here f(x) dx = x^3- 5.

Therefore F(x) = Int {f(x) dx = Int { x^3 - 5} dx +C.

F(x) = Int x^3 dx - 5 Int dx + C.

F(x) = (1/3+1))x^(3+1) - (5*x)/1 = (1/4)x^4 - 5x +C.

F(2) = (1/4)*2^4 - 5*2 + C = -6 + C

F(1) = (1/4)*1^4- 5*1 + C = 4- 10+ C = -19/4 +C

Therefore Int f(x) dx from x= 1 to x= 2 = F(2)- F(1) = -6 +19/4 = -5/4.

**Therefore Integral of (x^3-5) from x= 1 to x= 2 is -5/4**.