# What are values of a>0 if integral from 0 to a (3x^2+4x-5)dx=a^3-2?

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### 1 Answer

You need first to evaluate the definite integral, using the property of linearity of integral, such that:

`int_0^a (3x^2+4x-5) dx = int_0^a (3x^2)dx + int_0^a (4x)dx -` `int_0^a 5dx`

`int_0^a (3x^2+4x-5) dx = (3x^3/3 + 4x^2/2 - 5x)|_0^a`

Using the fundamental theorem of calculus yields:

`int_0^a (3x^2+4x-5) dx = a^3 - 0^3 + 2a^2 - 2*0^2 - 5a + 5*0`

`int_0^a (3x^2+4x-5) dx = a^3 + 2a^2 - 5a`

You need to equate the expressions `a^3 + 2a^2 - 5a` and `a^3 - 2 ` such that:

`a^3 + 2a^2 - 5a = a^3 - 2`

Reducing duplicate members yields:

`2a^2 - 5a + 2 = 0`

Using quadratic formula yields:

`a_(1,2) = (5+-sqrt(25 - 16))/4`

`a_(1,2) = (5+-sqrt9)/4 => a_(1,2) = (5+-3)/4`

`a_1 = 2 ; a_2 = 1/2`

**Hence, evaluating the positive values of a, under the given conditions, yields `a = 2 ; a = 1/2` .**