Answer:
[tex]z=1 \\ z=-7[/tex]
Step-by-step explanation:
Pretty straightforward question. We have the following definite integral:
[tex]$\int\limits^z_0 {\dfrac{1}{3}x - 1} \, dx $[/tex]
and this integral equals to [tex]\dfrac{7}{6}[/tex]. We want to find the value of [tex]z[/tex]
Considering the property that the integral of the sum of functions [tex]f[/tex] and [tex]g[/tex] equals the integral of [tex]f[/tex] plus the integral of [tex]g[/tex], we have
[tex]$\int\limits^z_0 {\dfrac{1}{3}x - 1} \, dx = \int _0^z\frac{1}{3}x \, dx-\int _0^z1 \,dx$[/tex]
Now we just have to solve each integral.
[tex]$\int _0^z\dfrac{1}{3}x \, dx = \dfrac{1}{3}\int _0^zx \, dx = \left[\dfrac{1}{3}\cdot \dfrac{x^2}{2}\right] \Big |_0^z = \dfrac{x^2}{6} \Big |_0^z = \dfrac{z^2}{6} - \dfrac{0^2}{6} =\boxed{ \dfrac{z^2}{6}}$[/tex]
Explanation: 1/3 is a constant, that's why we ended up calculating the integral on the given interval for [tex]x[/tex]. Then, for the integral of [tex]x[/tex] I just calculated the antiderivative, given as [tex]x^n = \dfrac{x^{n+1}}{n+1}[/tex] for [tex]n=1[/tex] and the Fundamental Theorem of Calculus.
For the other integral, we just have the definite integral of a constant.
[tex]$\int _0^z1 \,dx = 1(z-0) = \boxed{z}$[/tex]
Therefore,
[tex]$\int\limits^z_0 {\dfrac{1}{3}x - 1} \, dx = \dfrac{z^2}{6} + z$[/tex]
Now we can solve
[tex]$\dfrac{z^2}{6} + z = \dfrac{7}{6} \implies z^2+6z-7=0 \implies z = \frac{-6\pm \sqrt{6^2-4\cdot 1\cdot (-7)}}{2\cdot 1}$[/tex]
[tex]z=1 \\ z=-7[/tex]
