It is known that int_a^5 x^2+4x+1 dx = 132, what is the value of a.

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gsarora17 | (Level 2) Associate Educator

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`int_a^5(x^2+4x+1)dx=132`

`[x^3/3+4x^2/2+x]_a^5=132`

`[x^3/3+2x^2+x]_a^5=132`

`(5^3/3+2(5)^2+5)-(a^3/3+2a^2+a)=132`

`(125/3+50+5)-(a^3/3+2a^2+a)=132`

`((125+150+15)/3)-((a^3+6a^2+3a)/3)=132`

`((290-a^3-6a^2-3a)/3)=132`

`290-a^3-6a^2-3a=396`

`a^3+6a^2+3a+396-290=0`

`a^3+6a^2+3a+106=0`

Since the discriminant of this cubic equation is less than zero, so the equation has one real root and two complex conjugate roots.

Now to solve this cubic equation , let us depress the cubic equation by substituting a with (y-2).

So, `(y-2)^3+6(y-2)^2+3(y-2)+106=0`  

`(y^3-8-6y(y-2))+6(y^2+4-4y)+3y-6+106=0`

`y^3-8-6y^2+12y+6y^2+24-24y+3y-6+106=0`

`y^3-9y+116=0`

Now we have to solve this depressed cubic equation of the form y^3+Ay=B

`y=s-t , A=3st=-9 , B=s^3-t^3=-116`

`(A/(3t))^3-t^3=B`

`(-9/(3t))^3-t^3=-116`

`27/t^3+t^3=116`

`t^6+27=116t^3`

`t^6-116t^3+27=0`

Now to solve the above equation for t , let us reduce to the quadratic form by assuming , t^3=u.

So the equation reduces to,`u^2-116u+27=0`

 `u=(116+-sqrt(116^2-4*27))/2`

`u=(116+-sqrt(13456-108))/2`

`u=(116+-sqrt(13348))/2`

`u=(116+-sqrt(4*3337))/2 = 58+-sqrt(3337)`

`:.t=root(3)(58+-sqrt(3337))`

`s=-3/t`

`s=-3/root(3)(58+-sqrt(3337))`

`y=s-t and a=y-2`

` :.a=-2+s-t`

`a=-2-3/root(3)(58+sqrt(3337))-root(3)(58+sqrt(3337))`

` or a=-2-3/root(3)(58-sqrt(3337))-root(3)(58-sqrt(3337)) `

`or a=-7.48927`

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