evaluate the integral. (int_{0}^{1}3(1 + sqrt{x})^{7}dx) 28. -/1.11 points evaluate the integral. (use c for…

Explanation:

Step1: Use substitution for the first integral

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 $\int_{0}^{1}3(1+\sqrt{x})^7dx$ becomes $\int_{1}^{2}3u^7\cdot2(u - 1)du=\int_{1}^{2}(6u^8-6u^7)du$.

Step2: Integrate term - by - term

$\int(6u^8-6u^7)du=6\times\frac{u^{9}}{9}-6\times\frac{u^{8}}{8}+C=\frac{2}{3}u^{9}-\frac{3}{4}u^{8}+C$.

Step3: Evaluate the definite integral

$\left[\frac{2}{3}u^{9}-\frac{3}{4}u^{8}\right]_{1}^{2}=\left(\frac{2}{3}\times2^{9}-\frac{3}{4}\times2^{8}\right)-\left(\frac{2}{3}\times1^{9}-\frac{3}{4}\times1^{8}\right)=\left(\frac{2}{3}\times512-\frac{3}{4}\times256\right)-\left(\frac{2}{3}-\frac{3}{4}\right)=\left(\frac{1024}{3}-192\right)-\left(\frac{8 - 9}{12}\right)=\left(\frac{1024 - 576}{3}\right)+\frac{1}{12}=\frac{448}{3}+\frac{1}{12}=\frac{1792 + 1}{12}=\frac{1793}{12}$.

Step4: Use substitution for the second integral

Let $t=\sqrt{x}$, then $x = t^{2}$ and $dx = 2tdt$. The integral $\int25\sqrt{x}e^{\sqrt{x}}dx=\int25t\cdot e^{t}\cdot2tdt = 50\int t^{2}e^{t}dt$.

Step5: Use integration by parts

The formula for integration by parts is $\int u dv=uv-\int v du$. Let $u = t^{2}$, $dv=e^{t}dt$, then $du = 2tdt$, $v = e^{t}$. So $50\int t^{2}e^{t}dt=50\left(t^{2}e^{t}-2\int te^{t}dt\right)$.

Step6: Use integration by parts again for $\int te^{t}dt$

Let $u = t$, $dv=e^{t}dt$, then $du = dt$, $v = e^{t}$. So $\int te^{t}dt=te^{t}-\int e^{t}dt=te^{t}-e^{t}+C$.

Step7: Substitute back

$50\left(t^{2}e^{t}-2\left(te^{t}-e^{t}\right)\right)+C=50e^{t}(t^{2}-2t + 2)+C=50e^{\sqrt{x}}(x - 2\sqrt{x}+2)+C$.

Answer:

The value of $\int_{0}^{1}3(1+\sqrt{x})^7dx$ is $\frac{1793}{12}$ and the value of $\int25\sqrt{x}e^{\sqrt{x}}dx$ is $50e^{\sqrt{x}}(x - 2\sqrt{x}+2)+C$