Area under waveform in t-domain including 2 delta functions question:





my attempt:

$(a)$ $\displaystyle\int_{-\infty}^{\infty} f(t) dt=2 u (t)+r(t)-r(t-T)+Ku(t-T)$

then we draw it's diagram and found it was unbounded so,we doubt how area can be zero for any value of $K$ so, we need help here finding value of K (which we think  will come negative )

$(b)$ $\displaystyle\int_{0_{+}}^{\infty} f(t) dt=r(t)-r(t-T)+Ku(t-T)$

same diagram as part $(a)$ , and same story as above ,except  we pull above curve down by two units . 

so , we're unable to find value of $K$. any help will be appreciated 

thanking you !

edit : $r(t)$ is ramp function $= t u(t)$
please someone confirm if following  devlopment is right or  wrong .becuase answer of exercise question isn't provided in book "F F kuo"

we know 

$f(t)\xrightarrow{\mathcal{F}}F(j\omega); $ where, $F(j\omega) =\displaystyle\int_{-\infty}^{\infty} f(t) \ e^{-j\omega t} \ dt\implies F(0)=\text{area under time domain curve }$ 

So, $(a)\ F(j\omega)=2+T.\ Sa{(\omega T/2)}+Ke^{-j\omega T}\implies F(0)=2+T+K=\displaystyle\int_{-\infty}^{\infty}f(t)dt$

but area is given zero therefore , $K=-(2+T)$ . 

and it is resonable also becuase strength of impulse at $t=T$ must be equal to the negative summation of  area of regions preceding it. 

similarly,

for part $(b)\ $   $K=-T$ because time is starting from $t=0_{+}$ that's why excluding impulse of strength  $2$ at $t=0$

note :

$Sa (\omega T/2)=\dfrac{\sin\left(\dfrac{\omega T}{2}\right)}{\omega T/2}$