Question

How do you solve the following equation:


`\int_2^a (3x^2+x-1) dx=52` for
`a`?

Answer 1

`\displaystyle\\\int \limits_2^a(3x^2+x-1)\, dx=52\\\left|x^3+(x^2)/(2)-x\right|_2^a=52\\a^3+(a^2)/(2)-a-(2^3+(2^2)/(2)-2)=52\\a^3+(a^2)/(2)-a-(8+2-2)=52\\a^3+(a^2)/(2)-a-8=52\\a^3+(a^2)/(2)-a=60\\2a^3+a^2-2a=120\\2a^3+a^2-2a-120=0`

Now you need to solve the resulting equation, but it's not easy. The approx. solution is
`a\approx 3.8`

The exact solution is

`a=(1)/(6)\left(-1+\sqrt[3]{6461-78√(6861)}+\sqrt[3]{6461+78√(6861)}\right)`

Answer 2

a = 3.84

Explanation:

Let's integrate the function

3x^2 becomes 3x^3/3 =x^3

x becomes x^2/2

-1 becomes -x

The intergral is x^3 +x^2/2 -x

We take the upper limit minus the lower limit

a^3 +a^2/2 -a - (2^3 + 2^2/2 -2) and that is equal to 52

Simplify

a^3 + a^2/2 -a - (8+2-2) = 52

a^3 +a^2/2 -a -(8) = 52

Subtract 52 from each side

a^3 +a^2/2 -a -8-52 = 52-52

a^3 +a^2/2 -a -60 =0

Multiply by 2

2a^3 +a^2 -2a -120 = 0

Using a graphing system, we see it only has 1 real root

This is at approximately

3.84