Question

Need help question #2. Show steps please

Answer 1

Answer:

C

Step-by-step explanation:

We want to evaluate the definite integral:

$\displaystyle \int_0^1 (x+2)(3x^2+12x+1)^{1/2}\, dx$

Again, notice that the radicand is quite similar to the outside factor. So, we can use u-substitution again. We will let:

$\displaystyle u=3x^2+12x+1$

Then:

$\displaystyle \frac{du}{dx}=6x+12$

Hence:

$\displaystyle du=6x+12 \, dx$

And we can divide both sides by 6:

$\displaystyle \frac{1}{6}\, du=x+2\, dx$

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:

$u(0)=3(0)^2+12(0)+1=1$

And:

$u(1)=3(1)^2+12(1)+1=16$

Hence, our new limits of integration is from u = 1 to u = 16.

Perform the substitution:

$\displaystyle =\int_{1}^{16} u^{1/2}\Big(\frac{1}{6}\, du\Big)$

Simplify:

$\displaystyle =\frac{1}{6}\int_1^{16}u^{1/2}\, du$

Integrate:

$\displaystyle =\frac{1}{6}\Big(\frac{2}{3}u^{3/2}\Big)\Big|_{1}^{16}$

Simplify:

$=\displaystyle \frac{1}{9}\Big(u^{3/2}\Big|_1^{16}\Big)$

Evaluate:

$\displaystyle =\frac{1}{9}\Big(16^{3/2}-1^{3/2}\Big)=\frac{1}{9}(64-1)=7$

The answer is C.