Question

Calculus help ASAP!!!

Answer 1

Let $c\in(0,\infty)$. Then the definite integral can be split up at $t=c$ so that

$\displaystyle F(x)=\int_{t=2x}^{t=5x}\frac{\mathrm dt}t$
$\displaystyle=\int_{t=2x}^{t=c}\frac{\mathrm dt}t+\int_{t=c}^{t=5x}\frac{\mathrm dt}t$
$\displaystyle=-\int_{t=c}^{t=2x}\frac{\mathrm dt}t+\int_{t=c}^{t=5x}\frac{\mathrm dt}t$

Now take the derivative. By the fundamental theorem of calculus, you have

$\displaystyle\frac{\mathrm dF}{\mathrm dx}=\frac{\mathrm d}{\mathrm dx}\left[\int_{t=c}^{t=5x}\frac{\mathrm dt}t-\int_{t=c}^{t=2x}\frac{\mathrm dt}t\right]$
$=\displaystyle\frac1{5x}\dfrac{\mathrm d}{\mathrm dx}[5x]-\frac1{2x}\dfrac{\mathrm d}{\mathrm dx}[2x]$
$=\dfrac5{5x}-\dfrac2{2x}$
$=\dfrac1x-\dfrac1x$
$=0$

Then integrating with respect to $x$, we recover $F(x)$ and find that

$\displaystyle\int\dfrac{\mathrm dF}{\mathrm dx}\,\mathrm dx=\int0\,\mathrm dx$
$F(x)=C$

where $C$ is an arbitrary constant.