Question

hw. 6.4 integration by substitution

1. write the functions in the brackets such that the identity holds

(a) $x dx=d( )$
(b) $e^{x}dx = d( )$
(c) $n
eq - 1,x^{n}dx=d( )$
(d) $k
eq0,e^{kx + b}dx=d( )$
(e) $\frac{1}{x}dx=d( )$
(f) $\frac{1}{x - k}dx=d( )$

Explanation:

Step1: Recall derivative - integral relationship

The derivative of $y = f(x)$ is $dy=f'(x)dx$. We need to find $y$ such that $dy$ is the given expression on the left - hand side.

Step2: For (A)

The derivative of $\frac{1}{2}x^{2}$ using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $\frac{d}{dx}(\frac{1}{2}x^{2})=x$, so $xdx=d(\frac{1}{2}x^{2})$.

Step3: For (B)

The derivative of $e^{x}$ is $e^{x}$, so $e^{x}dx = d(e^{x})$.

Step4: For (C)

Using the power rule for integration $\int x^{n}dx=\frac{1}{n + 1}x^{n+1}+C$ ($n
eq - 1$), the derivative of $\frac{1}{n + 1}x^{n+1}$ is $x^{n}$, so $x^{n}dx=d(\frac{1}{n + 1}x^{n+1})$.

Step5: For (D)

Let $u=kx + b$, then $du=kdx$. The derivative of $\frac{1}{k}e^{kx + b}$ is $e^{kx + b}$, so $e^{kx + b}dx=d(\frac{1}{k}e^{kx + b})$.

Step6: For (E)

The derivative of $\ln|x|$ is $\frac{1}{x}$, so $\frac{1}{x}dx=d(\ln|x|)$.

Step7: For (F)

Let $u=x - k$, then $du = dx$. The derivative of $\ln|x - k|$ is $\frac{1}{x - k}$, so $\frac{1}{x - k}dx=d(\ln|x - k|)$.

Answer:

(A) $\frac{1}{2}x^{2}$
(B) $e^{x}$
(C) $\frac{1}{n + 1}x^{n+1}$
(D) $\frac{1}{k}e^{kx + b}$
(E) $\ln|x|$
(F) $\ln|x - k|$