There is no contradiction: just because a function goes to $\infty$ at $0$ does not mean its integral near $0$ should be infinite.
If you insist on interpreting the integral as area, here's a way to see that the area under the curve $y = 1/\sqrt{x}$ from $0$ to $1$ is finite. The area under the curve $y = 1/\sqrt{x}$ from $0$ to $1$ is the area of the region bounded by the $x$-axis, the $y$-axis, the line $x=1$, and $y = 1/\sqrt{x}$. But if we think about integrating in the $y$-direction, then this can also be interpreted as the area of the region bounded by $y=0$, $x = 1/y^2$, $x=0$, and $x=1$. This is just given by the integral
$$
1 + \int_1^\infty \frac{1}{y^2}~dy = 1 + 1 = 2.
$$
(The $1$ at the beginning comes from the square from $0$ to $1$. You should draw the region I described to see what I mean.)
Notice how I've computed the area of the same region, but I've managed to avoid any singularities.
In general this is a very common thing in calculus. There are other seemingly unintuitive results, such as a curve of infinite perimeter bounding a finite area (Koch snowflake) or a surface of revolution with infinite surface are bounding a region of finite volume (Gabriel's horn).
When you talk about integrals, you could be talking about one of many distinct, but related topics. You could mean Riemann integrals, indefinite integrals, or many other types of integral out there.
You're probably talking about the indefinite integral when you say that the integral of $\sin$ is $-\cos$. The thing about the indefinite integral is that it is not really a single function. That's why you often display it with the little ${}+C$ at the end. An indefinite integral is more like a set of functions, which are all equal under vertical translations.
So, $\int f$ could be considered the set of functions which have a derivative $f$. There are many functions which have the derivative $\sin$, and they are all of the form $-\cos+C$, where $C$ is some real number. This is why it is said the integral of $\sin$ is $-\cos$.
But why does the integral then not represent the area? Well, it does. What you were saying was that $-\cos(x)$ should be the area under $\sin(x)$ up to $x$. The question is, up to $x$ from where? There is no lower bound! If you pick the lower bound $0$, then
$$-\cos(x)-(-\cos(0))=-\cos(x)+1$$
You get that $+1$ you were wondering about, and it does represent the area under $\sin(x)$ from $0$ to $x$.
Best Answer
For exercise "a" you have to compute the following area:
I assume you know how to compute the area of a square and a triangle. Which is in the case of question "a" 4*2+0.5*4*4=16. Question b goes in a similar way.
Note that for question "c" you have to compute the area above the graph, since the graph itself lies underneath the x-axis: