Answer:
a) So, this integral is convergent.
b) So, this integral is divergent.
c) So, this integral is divergent.
Step-by-step explanation:
We calculate the next integrals:
a)
![\int_1^{\infty} e^{-2x} dx=\left[-\frac{e^{-2x}}{2}\right]_1^{\infty}\\\\\int_1^{\infty} e^{-2x} dx=-\frac{e^{-\infty}}{2}+\frac{e^{-2}}{2}\\\\\int_1^{\infty} e^{-2x} dx=\frac{e^{-2}}{2}\\](https://tex.z-dn.net/?f=%5Cint_1%5E%7B%5Cinfty%7D%20e%5E%7B-2x%7D%20dx%3D%5Cleft%5B-%5Cfrac%7Be%5E%7B-2x%7D%7D%7B2%7D%5Cright%5D_1%5E%7B%5Cinfty%7D%5C%5C%5C%5C%5Cint_1%5E%7B%5Cinfty%7D%20e%5E%7B-2x%7D%20dx%3D-%5Cfrac%7Be%5E%7B-%5Cinfty%7D%7D%7B2%7D%2B%5Cfrac%7Be%5E%7B-2%7D%7D%7B2%7D%5C%5C%5C%5C%5Cint_1%5E%7B%5Cinfty%7D%20e%5E%7B-2x%7D%20dx%3D%5Cfrac%7Be%5E%7B-2%7D%7D%7B2%7D%5C%5C)
So, this integral is convergent.
b)
![\int_1^{2}\frac{dz}{(z-1)^2}=\left[-\frac{1}{z-1}\right]_1^2\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\frac{1}{1-1}+\frac{1}{2-1}\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\infty\\](https://tex.z-dn.net/?f=%5Cint_1%5E%7B2%7D%5Cfrac%7Bdz%7D%7B%28z-1%29%5E2%7D%3D%5Cleft%5B-%5Cfrac%7B1%7D%7Bz-1%7D%5Cright%5D_1%5E2%5C%5C%5C%5C%5Cint_1%5E%7B2%7D%5Cfrac%7Bdz%7D%7B%28z-1%29%5E2%7D%3D-%5Cfrac%7B1%7D%7B1-1%7D%2B%5Cfrac%7B1%7D%7B2-1%7D%5C%5C%5C%5C%5Cint_1%5E%7B2%7D%5Cfrac%7Bdz%7D%7B%28z-1%29%5E2%7D%3D-%5Cinfty%5C%5C)
So, this integral is divergent.
c)
![\int_1^{\infty} \frac{dx}{\sqrt{x}}=\left[2\sqrt{x}\right]_1^{\infty}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=2\sqrt{\infty}-2\sqrt{1}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=\infty\\](https://tex.z-dn.net/?f=%5Cint_1%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdx%7D%7B%5Csqrt%7Bx%7D%7D%3D%5Cleft%5B2%5Csqrt%7Bx%7D%5Cright%5D_1%5E%7B%5Cinfty%7D%5C%5C%5C%5C%5Cint_1%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdx%7D%7B%5Csqrt%7Bx%7D%7D%3D2%5Csqrt%7B%5Cinfty%7D-2%5Csqrt%7B1%7D%5C%5C%5C%5C%5Cint_1%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdx%7D%7B%5Csqrt%7Bx%7D%7D%3D%5Cinfty%5C%5C)
So, this integral is divergent.
"T<u>he function is decreasing":</u>
This means the slope of the function will be negative at the given point. Anywhere the line is going down, from left to right, is decreasing. This eliminates choices A and C.
<u>"The function is concave down":</u>
This means the curve of the function will be open down (think of an upside down u). This eliminates choice E.
<u>"f(x)>0":</u>
This is anywhere the curve is above the x axis. This eliminates choice D.
The answer is Choice B) x=2.
Hope this helps!!
The method that is not correct is brandon's method.
General equation: y = mx + c
where m is the slope.
So if there’s a negative sign on mx, it is a negative slope and vice versa.
