Question

$$\sum_{x=3}^{75}\int4x^{3}-2x^{2}+5in\ 30-\cos\ 180+C$$

Answer

$$s=(24.330671508534*10^35*nt*IM*x^3-2*x^2+150*IM*n-cos(180)+C)/(u*m*x)$$

Solution


Simplify  \({3}^{75}\)  to  \(6.082668\times {10}^{35}\).
\[sumx=6.082668\times {10}^{35}\imath nt\times 4{x}^{3}-2{x}^{2}+5\imath n\times 30-\cos{180}+C\]
Simplify  \(6.082668\times {10}^{35}\imath nt\times 4{x}^{3}\)  to  \((24.330672){x}^{3}\times {10}^{35}\imath nt\).
\[sumx=24.330672{x}^{3}\times {10}^{35}\imath nt-2{x}^{2}+5\imath n\times 30-\cos{180}+C\]
Regroup terms.
\[sumx=24.330672\times {10}^{35}nt\imath {x}^{3}-2{x}^{2}+5\imath n\times 30-\cos{180}+C\]
Simplify  \(5\imath n\times 30\)  to  \(150n\imath \).
\[sumx=24.330672\times {10}^{35}nt\imath {x}^{3}-2{x}^{2}+150n\imath -\cos{180}+C\]
Regroup terms.
\[sumx=24.330672\times {10}^{35}nt\imath {x}^{3}-2{x}^{2}+150\imath n-\cos{180}+C\]
Divide both sides by \(u\).
\[smx=\frac{24.330672\times {10}^{35}nt\imath {x}^{3}-2{x}^{2}+150\imath n-\cos{180}+C}{u}\]
Divide both sides by \(m\).
\[sx=\frac{\frac{24.330672\times {10}^{35}nt\imath {x}^{3}-2{x}^{2}+150\imath n-\cos{180}+C}{u}}{m}\]
Simplify  \(\frac{\frac{24.330672\times {10}^{35}nt\imath {x}^{3}-2{x}^{2}+150\imath n-\cos{180}+C}{u}}{m}\)  to  \(\frac{24.330672\times {10}^{35}nt\imath {x}^{3}-2{x}^{2}+150\imath n-\cos{180}+C}{um}\).
\[sx=\frac{24.330672\times {10}^{35}nt\imath {x}^{3}-2{x}^{2}+150\imath n-\cos{180}+C}{um}\]
Divide both sides by \(x\).
\[s=\frac{\frac{24.330672\times {10}^{35}nt\imath {x}^{3}-2{x}^{2}+150\imath n-\cos{180}+C}{um}}{x}\]
Simplify  \(\frac{\frac{24.330672\times {10}^{35}nt\imath {x}^{3}-2{x}^{2}+150\imath n-\cos{180}+C}{um}}{x}\)  to  \(\frac{24.330672\times {10}^{35}nt\imath {x}^{3}-2{x}^{2}+150\imath n-\cos{180}+C}{umx}\).
\[s=\frac{24.330672\times {10}^{35}nt\imath {x}^{3}-2{x}^{2}+150\imath n-\cos{180}+C}{umx}\]