in exercises 25 - 29, evaluate the integral.\n25. $int_{0}^{1}\frac{36}{(2x + 1)^{3}}dx$\n26…

25.

Explanation:

Step1: Use substitution

Let $u = 2x+1$, then $du=2dx$. When $x = 0$, $u = 1$; when $x = 1$, $u=3$. And $dx=\frac{1}{2}du$. The integral becomes $\int_{0}^{1}\frac{36}{(2x + 1)^{3}}dx=36\times\frac{1}{2}\int_{1}^{3}u^{-3}du = 18\int_{1}^{3}u^{-3}du$.

Step2: Apply power - rule for integration

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$. So, $18\int_{1}^{3}u^{-3}du=18\times\left[\frac{u^{-3 + 1}}{-3+1}\right]{1}^{3}=18\times\left[-\frac{1}{2u^{2}}\right]{1}^{3}$.

Step3: Evaluate the definite integral

$18\times\left(-\frac{1}{2u^{2}}\right)\big|_{1}^{3}=18\times\left(-\frac{1}{18}+\frac{1}{2}\right)=18\times\left(\frac{-1 + 9}{18}\right)=8$.

Answer:

$8$

26.

Explanation:

Step1: Split the integral

$\int_{1}^{2}\left(x+\frac{1}{x^{2}}\right)dx=\int_{1}^{2}x dx+\int_{1}^{2}x^{-2}dx$.

Step2: Apply power - rule for integration

For $\int_{1}^{2}x dx=\left[\frac{x^{2}}{2}\right]{1}^{2}=\frac{4}{2}-\frac{1}{2}=\frac{3}{2}$, and $\int{1}^{2}x^{-2}dx=\left[\frac{x^{-2 + 1}}{-2 + 1}\right]{1}^{2}=\left[-\frac{1}{x}\right]{1}^{2}=-\frac{1}{2}+1=\frac{1}{2}$.

Step3: Sum the results of the two integrals

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

Answer:

$2$

27.

Explanation:

Step1: Recall the antiderivative of $\sec x\tan x$

The antiderivative of $\sec x\tan x$ is $\sec x$. So, $\int_{-\frac{\pi}{3}}^{0}\sec x\tan xdx=[\sec x]_{-\frac{\pi}{3}}^{0}$.

Step2: Evaluate the definite integral

$[\sec x]_{-\frac{\pi}{3}}^{0}=\sec(0)-\sec\left(-\frac{\pi}{3}\right)=1 - 2=-1$.

Answer:

$-1$

28.

Explanation:

Step1: Use substitution

Let $u = 1 - x^{2}$, then $du=-2xdx$. When $x=-1$, $u = 0$; when $x = 1$, $u = 0$. So, $\int_{-1}^{1}2x\sin(1 - x^{2})dx=-\int_{0}^{0}\sin(u)du$.

Step2: Evaluate the definite integral

Since $\int_{a}^{a}f(x)dx = 0$ for any real - valued function $f(x)$ and real number $a$, $-\int_{0}^{0}\sin(u)du = 0$.

Answer:

$0$

29.

Explanation:

Step1: Recall the antiderivative of $\frac{1}{y + 1}$

The antiderivative of $\frac{1}{y + 1}$ is $\ln|y + 1|$. So, $\int_{0}^{2}\frac{2}{y + 1}dy=2\int_{0}^{2}\frac{1}{y + 1}dy$.

Step2: Apply the fundamental theorem of calculus

$2\int_{0}^{2}\frac{1}{y + 1}dy=2[\ln(y + 1)]_{0}^{2}=2(\ln3-\ln1)=2\ln3$.

Answer:

$2\ln3$