Question

in exercises 67–70, find ( f(x) ).

1. ( f(x) = int_{1}^{x} \frac{1}{t} dt )
2. ( f(x) = int_{0}^{x} \tan t dt )
3. ( f(x) = int_{1}^{3x} \frac{1}{t} dt )
4. ( f(x) = int_{1}^{x^2} \frac{1}{t} dt )

Explanation:

Step1: Apply First Fundamental Theorem of Calculus (FTC1)

For $F(x)=\int_{a}^{x}f(t)dt$, $F'(x)=f(x)$.
For 67: $F(x)=\int_{1}^{x}\frac{1}{t}dt$, so $F'(x)=\frac{1}{x}$.

Step2: Apply FTC1 directly

For 68: $F(x)=\int_{0}^{x}\tan t dt$, so $F'(x)=\tan x$.

Step3: Apply FTC1 + Chain Rule

For $F(x)=\int_{a}^{u(x)}f(t)dt$, $F'(x)=f(u(x))\cdot u'(x)$.
For 69: $u(x)=3x$, $u'(x)=3$, $f(t)=\frac{1}{t}$.
$F'(x)=\frac{1}{3x}\cdot3=\frac{1}{x}$.

Step4: Apply FTC1 + Chain Rule

For 70: $u(x)=x^2$, $u'(x)=2x$, $f(t)=\frac{1}{t}$.
$F'(x)=\frac{1}{x^2}\cdot2x=\frac{2}{x}$.

Answer:

1. $F'(x)=\frac{1}{x}$
2. $F'(x)=\tan x$
3. $F'(x)=\frac{1}{x}$
4. $F'(x)=\frac{2}{x}$