Question

QUIZ MATH!!!

Can y'all solve this ?
Good luck ​

Answer 1

Answer:

69.72

Step-by-step explanation:

$\int\limits^4_1( {x\sqrt{x} + \frac{1}{x\sqrt{x}} } )^{2} \, dx$

First, $\int\ ({x\sqrt{x} + \frac{1}{x\sqrt{x}} } )^{2} \, dx$

= $\int\ (x^{3} + \frac{1}{x^{3}} + 2) \, dx$

=$\int\ x^{3} \, dx + \int\ {x^{-3}} \, dx + \int\ 2 \, dx$

=$\frac{x^{4} }{4} + \frac{x^{-2} }{-2} + 2x$

=$\frac{x^{4} }{4} + \frac{-1}{2x^{2} } + 2x$

Now,

$\int\limits^4_1( {x\sqrt{x} + \frac{1}{x\sqrt{x}} } )^{2} \, dx$      = [ $\frac{x^{4} }{4} + \frac{-1}{2x^{2} } + 2x$]⁴₁

= {(4⁴ - 1⁴)÷4} + {-1÷2(4²-1²)} + 2(4-1)

= $\frac{255}{4} - \frac{1}{30} + 6$

= $\frac{3825 - 2 + 360}{60}$

=$\frac{4183}{60}$

≈69.72