91Ó°ÊÓ

We integrate the given function with respect to z and get $$\int_{0}^{2-x} 4yz dz = 4y\int_{0}^{2-x} z dz.$$ Then, we find the integral of z with respect to z: $$4y\int_{0}^{2-x} z dz = 4y[\frac{1}{2}z^2]_{0}^{2-x} = 4y[\frac{1}{2}(2-x)^2-\frac{1}{2}(0)^2].$$ Now we continue integrating with respect to y:

We plug the result from step 1 and integrate with respect to y: $$\int_{0}^{\sqrt{1-x^{2}}} 4y[\frac{1}{2}(2-x)^2] dy = 2\int_{0}^{\sqrt{1-x^{2}}} y(2-x)^2 dy.$$ Now, we integrate y with respect to y: $$2\int_{0}^{\sqrt{1-x^{2}}} y(2-x)^2 dy = 2[\frac{1}{2}y^2(2-x)^2]_{0}^{\sqrt{1-x^{2}}} = [\sqrt{1-x^{2}}(2-x)^2].$$ Finally, we integrate with respect to x.

We integrate the result from step 2 with respect to x: $$\int_{0}^{1} [\sqrt{1-x^{2}}(2-x)^2]dx.$$ To integrate this function, we can use u-substitution. We let \(u = x^2\), so \(\frac{du}{dx} = 2x\). "@/analysis/Temperature_Humidity_Sensor_Humidity_Store_Data_Report_State_Check/PredictBackwardInOut1/$/" Then, we can rewrite the integral as follows: $$\int_{0}^{1} [\sqrt{1-u}(2-\sqrt{u})^2] \frac{1}{2\sqrt{u}} du.$$ Now, we can simplify the integral and compute it: $$\frac{1}{2}\int_{0}^{1} (1-u)(4-4\sqrt{u}+u) du.$$ Integrating term by term, we have $$\frac{1}{2}\left[\int_{0}^{1} 4 - 4\sqrt{u}+u -4u + 4u\sqrt{u}-u^2 du\right].$$ Now, we compute each of these integrals individually: \begin{align*} \int_{0}^{1} 4 du &= 4u\Big|_{0}^{1}=4, \\ \int_{0}^{1} - 4\sqrt{u} du &= -\frac{8}{3}u^{3/2}\Big|_{0}^{1}=-\frac{8}{3}, \\ \int_{0}^{1} u du &= \frac{1}{2}u^2\Big|_{0}^{1}=\frac{1}{2}, \\ \int_{0}^{1} -4u du &= -2u^2\Big|_{0}^{1}=-2, \\ \int_{0}^{1} 4u\sqrt{u} du &= \frac{32}{15}u^{5/2}\Big|_{0}^{1}=\frac{32}{15}, \\ \int_{0}^{1} -u^2 du &= -\frac{1}{3}u^3\Big|_{0}^{1}=-\frac{1}{3}. \end{align*}

Now, we sum the results of the previous step and multiply them by the constant factor: $$\frac{1}{2}\left[4 - \frac{8}{3} + \frac{1}{2} - 2 + \frac{32}{15} - \frac{1}{3}\right] = \frac{1}{2}\left[\frac{87}{30}\right] = \frac{87}{60}.$$ Thus, the integral evaluates to \(\frac{87}{60}\).