27. - / 1.07 points evaluate the integral. $int_{0}^{1}3(1 + sqrt{x})^{7}dx$ resources watch it

Explanation:

Step1: Use substitution

Let $u = 1+\sqrt{x}$, then $x=(u - 1)^2$ and $dx = 2(u - 1)du$. When $x = 0$, $u=1$; when $x = 1$, $u=2$. The integral $\int_{0}^{1}3(1 + \sqrt{x})^{7}dx$ becomes $\int_{1}^{2}3u^{7}\cdot2(u - 1)du=6\int_{1}^{2}(u^{8}-u^{7})du$.

Step2: Integrate term - by - term

We know that $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n\neq - 1$). So, $6\int_{1}^{2}(u^{8}-u^{7})du=6\left[\frac{u^{9}}{9}-\frac{u^{8}}{8}\right]_{1}^{2}$.

Step3: Evaluate the definite integral

First, substitute $u = 2$: $6\left(\frac{2^{9}}{9}-\frac{2^{8}}{8}\right)=6\left(\frac{512}{9}-32\right)=6\left(\frac{512 - 288}{9}\right)=6\times\frac{224}{9}=\frac{448}{3}$. Then substitute $u = 1$: $6\left(\frac{1}{9}-\frac{1}{8}\right)=6\times\frac{8 - 9}{72}=-\frac{1}{12}$. Subtract the two results: $\frac{448}{3}-\left(-\frac{1}{12}\right)=\frac{1792 + 1}{12}=\frac{1793}{12}$.

Answer:

$\frac{1793}{12}$