How do you find the integral of #20+ (4s^4)/sqrts ds#?

Question

HIX Tutor · Accepted Answer

undefined To find the integral of 20 + (4s^4)/sqrt(s) ds, you can follow these steps: 
1. Rewrite the expression as 20 + 4s^3 * s / sqrt(s). 
2. Take the integral of each term separately: 
- The integral of 20 is 20s. 
- The integral of 4s^3 * s / sqrt(s) can be simplified to 4s^4 * sqrt(s). 
3. Therefore, the integral of 20 + (4s^4)/sqrt(s) ds is 20s + 4s^4 * sqrt(s) + C, where C is the constant of integration.

Charles Arocha · Answer

#20s+8/9s^(9/2)+C#

We have

#int(20+(4s^4)/sqrts)ds#

Divide this up because addition can be used to separate integrals:

#=int20ds+int(4s^4)/sqrtsds#

Here's how the second integrand can be made simpler:

#(4s^4)/sqrts=(4s^4)/s^(1/2)=4s^(4-1/2)=4s^(7/2)#

Thus, we possess the integral

#int20ds+4ints^(7/2)ds#

The first integral is just #20s+C#, and find the next integral using the rule #ints^nds=s^(n+1)/(n+1)+C#.

#=20s+4(s^(7/2+1)/(7/2+1))+C#

#=20s+4(2/9)s^(9/2)+C#

#=20s+8/9s^(9/2)+C#

Evelyn Agard · Answer

undefined To find the integral of $20 + \frac{4s^4}{\sqrt{s}}$ with respect to $s$, you can break it down into two separate integrals: one for $20$ and another for $\frac{4s^4}{\sqrt{s}}$.

The integral of $20$ with respect to $s$ is $20s$.

To find the integral of $\frac{4s^4}{\sqrt{s}}$, you can rewrite it as $4s^4 \cdot s^{-1/2}$. Then, using the power rule for integration, the integral of $s^n$ with respect to $s$ is $\frac{s^{n+1}}{n+1}$, so the integral of $4s^4 \cdot s^{-1/2}$ is $\frac{4s^{4+\frac{1}{2}}}{4+\frac{1}{2}}$.

This simplifies to $\frac{4s^{9/2}}{9/2} = \frac{8s^{9/2}}{9}$.

Thus, the integral of $20 + \frac{4s^4}{\sqrt{s}}$ with respect to $s$ is $20s + \frac{8s^{9/2}}{9}$ plus the constant of integration, $C$.