Find the value of `int_(-1)^2 4/(x^3) dx`

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The value of `int_(-1)^2 4/x^3 dx` has to be determined.

`int 4/x^3 dx = 4*x^-2/-2 = -2/x^2`

Between the limits -1 and 2 the value is:

`-2(1/4 - 1/1) `

=> `-2(-3/4)`

=> ` 3/2`

The value of `int_(-1)^2 4/x^3 dx = 3/2`

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

The integral `int_(-1)^2 4/x^3 dx` has to be evaluated.

The integral `int x^n dx` is `x^(n+1)/(n+1)`

`int_(-1)^2 4/x^3 dx`

use the property `1/x^n = x^(-n)`

= `int_(-1)^2 4*x^-3 dx`

Now as 4 is a constant it can be taken out of the integral.

= `4*int_(-1)^2 x^-3 dx`

= `[4*x^(-3+1)/(-3+1)]_-1^2`

= `[4*x^(-2)/(-2)]_-1^2`

= `[-2*x^(-2)]_-1^2`

= `-2*(2^-2 - (-1)^-2)`

= `-2*(1/4 - 1)`

= `-2*(-3/4)`

= `3/2`

The integral `int_(-1)^2 4/x^3 dx = 3/2`

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