Answer:
C
Step-by-step explanation:
We want to evaluate the definite integral:
[tex]\displaystyle \int_0^1 (x+2)(3x^2+12x+1)^{1/2}\, dx[/tex]
Again, notice that the radicand is quite similar to the outside factor. So, we can use u-substitution again. We will let:
[tex]\displaystyle u=3x^2+12x+1[/tex]
Then:
[tex]\displaystyle \frac{du}{dx}=6x+12[/tex]
Hence:
[tex]\displaystyle du=6x+12 \, dx[/tex]
And we can divide both sides by 6:
[tex]\displaystyle \frac{1}{6}\, du=x+2\, dx[/tex]
Note that the limits of integration of our original integral (from x = 0 to x = 1) is in the domain of x. Since we changed variables, we should also change the limits of integration to u. So:
[tex]u(0)=3(0)^2+12(0)+1=1[/tex]
And:
[tex]u(1)=3(1)^2+12(1)+1=16[/tex]
Hence, our new limits of integration is from u = 1 to u = 16.
Perform the substitution:
[tex]\displaystyle =\int_{1}^{16} u^{1/2}\Big(\frac{1}{6}\, du\Big)[/tex]
Simplify:
[tex]\displaystyle =\frac{1}{6}\int_1^{16}u^{1/2}\, du[/tex]
Integrate:
[tex]\displaystyle =\frac{1}{6}\Big(\frac{2}{3}u^{3/2}\Big)\Big|_{1}^{16}[/tex]
Simplify:
[tex]=\displaystyle \frac{1}{9}\Big(u^{3/2}\Big|_1^{16}\Big)[/tex]
Evaluate:
[tex]\displaystyle =\frac{1}{9}\Big(16^{3/2}-1^{3/2}\Big)=\frac{1}{9}(64-1)=7[/tex]
The answer is C.
