How do you find the definite integral for: #[(6x^2+2) / sqrt(x)] dx # for the intervals #[1, 5]#?

Question

Emily Ammerman · Accepted Answer

#~~136.71# #"rewrite as"# #(6x^2)/(x^(1/2))+2/x^(1/2)=6x^(3/2)+2x^(-1/2)# #rArrint_1^5(6x^(3/2)+2x^(-1/2))dx# #"integrate each term using the "color(blue)"power rule"# #• int(ax^n)=a/(n+1)x^(n+1); n!=-1# #=[12/5x^(5/2)+4x^(1/2)]_1^5# #=(12/5. 5^(5/2)+4. 5^(1/2))-(12/5 .1+4.1)# #=60sqrt5+4sqrt5-32/5# #=64sqrt5-32/5~~136.71" to 2 dec. places"#

Emma Aldridge · Answer

# 64sqrt5-6.4#

Let, #I=int_1^5 (6x^2+2)/sqrtx dx.#

We know that, #intx^ndx=x^(n+1)/(n+1), n!=-1.#

#:. I=int_1^5(6x^2+2)/sqrtxdx,#

#=int_1^5(6x^2+2)*x^(-1/2)dx,#

#=int_1^5{(6x^(2-1/2)+2x^(-1/2)}dx,#

#=6int_1^5x^(3/2)dx+2int_1^5x^(-1/2)dx,#

#=6[x^(3/2+1)/(3/2+1)]_1^5+2[x^(-1/2+1)/(-1/2+1)]_1^5,#

#=6*2/5[x^(5/2)]_1^5+2*2[x^(1/2)]_1^5,#

#=12/5[5^(5/2)-1^(5/2)]+4[5^(1/2)-1^(1/2)]#

#=12/5(25sqrt5-1)+4(sqrt5-1),#

#=60sqrt5-12/5+4sqrt5-4,#

# rArr I=64sqrt5-6.4#

Enjoy Maths.!

Luke Alvarez · Answer

undefined To find the definite integral of $ \frac{6x^2 + 2}{\sqrt{x}} $ over the interval $[1, 5]$, you can use the Fundamental Theorem of Calculus. First, rewrite the expression as $ 6x^2x^{-1/2} + 2x^{-1/2} $. Then integrate each term separately:

$$ \int_{1}^{5} \frac{6x^2 + 2}{\sqrt{x}} \, dx = \int_{1}^{5} 6x^{2}x^{-1/2} \, dx + \int_{1}^{5} 2x^{-1/2} \, dx $$

$$ = \left[ 6 \cdot \frac{x^{2 + 1}}{2 + 1} \right]_{1}^{5} + \left[ 2 \cdot \frac{x^{1/2 + 1}}{1/2 + 1} \right]_{1}^{5} $$

$$ = \left[ 2x^{5/2} \right]_{1}^{5} + \left[ 4x^{3/2} \right]_{1}^{5} $$

$$ = 2(5^{5/2} - 1) + 4(5^{3/2} - 1) $$

$$ \approx 101.656 $$

Therefore, the definite integral of $ \frac{6x^2 + 2}{\sqrt{x}} $ over the interval $[1, 5]$ is approximately $ 101.656 $.