Question

My friend needs help with this but it's above my math level, pls help

Answer 1

`z=1 \\ z=-7`

Explanation:

Pretty straightforward question. We have the following definite integral:

`$\int\limits^z_0 {(1)/(3)x - 1} \, dx $`

and this integral equals to
`(7)/(6)`. We want to find the value of
`z`

Considering the property that the integral of the sum of functions
`f` and
`g` equals the integral of
`f` plus the integral of
`g`, we have

`$\int\limits^z_0 {(1)/(3)x - 1} \, dx = \int _0^z(1)/(3)x \, dx-\int _0^z1 \,dx$`

Now we just have to solve each integral.

`$\int _0^z(1)/(3)x \, dx = (1)/(3)\int _0^zx \, dx = \left[(1)/(3)\cdot (x^2)/(2)\right] \Big |_0^z = (x^2)/(6) \Big |_0^z = (z^2)/(6) - (0^2)/(6) =\boxed{ (z^2)/(6)}$`

Explanation: 1/3 is a constant, that's why we ended up calculating the integral on the given interval for
`x`. Then, for the integral of
`x` I just calculated the antiderivative, given as
`x^n = (x^(n+1))/(n+1)` for
`n=1` and the Fundamental Theorem of Calculus.

For the other integral, we just have the definite integral of a constant.

`$\int _0^z1 \,dx = 1(z-0) = \boxed{z}$`

Therefore,

`$\int\limits^z_0 {(1)/(3)x - 1} \, dx = (z^2)/(6) + z$`

Now we can solve

`$(z^2)/(6) + z = (7)/(6) \implies z^2+6z-7=0 \implies z = (-6\pm √(6^2-4\cdot 1\cdot (-7)))/(2\cdot 1)$`

`z=1 \\ z=-7`