integrate the following definite integral\n1. $int_{2}^{3}\frac{x^{3}-5x^{2}}{x}dx$\n2. $int_{0}^{\frac{pi}{4…

1. $\int_{2}^{3}\frac{x^{3}-5x^{2}}{x}dx$

Explanation:

Step1: Simplify the integrand

$\frac{x^{3}-5x^{2}}{x}=x^{2}-5x$

Step2: Integrate term - by - term

$\int(x^{2}-5x)dx=\frac{x^{3}}{3}-\frac{5x^{2}}{2}+C$

Step3: Apply the fundamental theorem of calculus

$\left[\frac{x^{3}}{3}-\frac{5x^{2}}{2}\right]_{2}^{3}=\left(\frac{3^{3}}{3}-\frac{5\times3^{2}}{2}\right)-\left(\frac{2^{3}}{3}-\frac{5\times2^{2}}{2}\right)$ $=(9 - \frac{45}{2})-(\frac{8}{3}-10)$ $=9-\frac{45}{2}-\frac{8}{3}+10$ $=19-\frac{135 + 16}{6}=19-\frac{151}{6}=\frac{114 - 151}{6}=-\frac{37}{6}$

2. $\int_{0}^{\frac{\pi}{4}}\sin(2x)dx$

Explanation:

Step1: Use substitution

Let $u = 2x$, then $du=2dx$ and $dx=\frac{1}{2}du$. When $x = 0$, $u = 0$; when $x=\frac{\pi}{4}$, $u=\frac{\pi}{2}$. $\int\sin(2x)dx=\frac{1}{2}\int\sin(u)du$

Step2: Integrate $\sin(u)$

$\frac{1}{2}\int\sin(u)du=-\frac{1}{2}\cos(u)+C=-\frac{1}{2}\cos(2x)+C$

Step3: Apply the fundamental theorem of calculus

$\left[-\frac{1}{2}\cos(2x)\right]_{0}^{\frac{\pi}{4}}=-\frac{1}{2}\cos\left(\frac{\pi}{2}\right)+\frac{1}{2}\cos(0)$ $=0+\frac{1}{2}=\frac{1}{2}$

3. $\int_{0}^{1}x^{2}(x^{3}+5)^{2}dx$

Explanation:

Step1: Use substitution

Let $u=x^{3}+5$, then $du = 3x^{2}dx$ and $x^{2}dx=\frac{1}{3}du$. When $x = 0$, $u = 5$; when $x = 1$, $u=6$. $\int x^{2}(x^{3}+5)^{2}dx=\frac{1}{3}\int u^{2}du$

Step2: Integrate $u^{2}$

$\frac{1}{3}\int u^{2}du=\frac{1}{3}\times\frac{u^{3}}{3}+C=\frac{1}{9}(x^{3}+5)^{3}+C$

Step3: Apply the fundamental theorem of calculus

$\left[\frac{1}{9}(x^{3}+5)^{3}\right]_{0}^{1}=\frac{1}{9}(6^{3}-5^{3})$ $=\frac{1}{9}(216 - 125)=\frac{91}{9}$

4. $\int_{1}^{2}4xe^{x^{2}}dx$

Explanation:

Step1: Use substitution

Let $u=x^{2}$, then $du = 2xdx$ and $4xe^{x^{2}}dx=2e^{u}du$. When $x = 1$, $u = 1$; when $x = 2$, $u = 4$. $\int4xe^{x^{2}}dx=2\int e^{u}du$

Step2: Integrate $e^{u}$

$2\int e^{u}du=2e^{u}+C=2e^{x^{2}}+C$

Step3: Apply the fundamental theorem of calculus

$\left[2e^{x^{2}}\right]_{1}^{2}=2e^{4}-2e^{1}=2e(e^{3}-1)$

5. $\int_{0}^{1}xe^{x}dx$

Explanation:

Step1: Use integration by parts

The formula for integration by parts is $\int u dv=uv-\int v du$. Let $u = x$ and $dv=e^{x}dx$. Then $du = dx$ and $v = e^{x}$. $\int xe^{x}dx=xe^{x}-\int e^{x}dx$

Step2: Integrate $e^{x}$

$xe^{x}-\int e^{x}dx=xe^{x}-e^{x}+C=(x - 1)e^{x}+C$

Step3: Apply the fundamental theorem of calculus

$\left[(x - 1)e^{x}\right]_{0}^{1}=(1 - 1)e^{1}-(0 - 1)e^{0}=1$

Answer:

1. $-\frac{37}{6}$
2. $\frac{1}{2}$
3. $\frac{91}{9}$
4. $2e(e^{3}-1)$
5. $1$