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How does one answer these type of question:

(a) If f is continuous, decreasing function on [1, $\infty$) and $\lim\limits_{x\to\infty} f(x) = 0$, then $\int_{1}^{\infty} f(x) dx$ is convergent.

(b) If $f(x) \leq g(x)$ for all x and $\int_{0}^{\infty} f(x)dx$ diverges, then $\int_{0}^{\infty} g(x)dx$ also diverges.

(c) If $\int_{7}^{\infty}f(x)dx$ convereges and let $c \in R$, then $\int_{7}^{\infty} cf(x)dx$ also converges.

For c) i just argued that the constant doesn't matter because of the integral property of putting the constant out. So if it diverges or converges you just multiply the constant. So true. Bad explanation i think

(d) Let $a \in R$. If $0 \leq f(x) \leq g(x)$ for all $x \in [a,\infty)$ and $\int_{a}^{\infty} g(x)dx$ diverges, then $\int_{a}^{\infty} f(x)dx$ also diverges.

i have no idea for d.

I was reading theorems in this section in my textbook but none of them really helped me figure this out.

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    $(1)$ is incorrect. Take $f(x)=\frac{1}{x}$. – 2017-02-16

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(a) is false. Take for example $f:x\mapsto\frac 1x$

(b) is also false (and would be true with the additional assumption that $f(x)\ge0$ for all $x\ge0$). Consider for example $f:x\mapsto-1$ and $g:x\mapsto\exp(-x)$.

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    so 1/x is a p-series that diverges okay I get that. for b) $f(x) \leq g(x)$ I don't know what this even does to the integral? Could you explain? – 2017-02-16
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    @user349557: Could explain the expression "p-series" ? I don't see any series around ... For b), I don't know what to explain : it should be clear that for all $x\ge0$ we have $f(x)=-1<\exp(-x)=g(x)$, that $\int_0^{+\infty}f(x)\,dx$ diverges and that $\int_0^{+\infty}\exp(-x)\,dx$ converges. – 2017-02-16
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    Its just the denominator has the power value 1, if p < 1 it diverges if p > 1 it coverges. Okay so I kind of get b, if the int was 1 to infty instead of 0 to infty would this statement hold True? I actually used the word p series cus i read something on it while googling ignore that. "Integral of power functions" is a better word – 2017-02-16
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    This counterexample works exactly the same on any interval $[a,+\infty)$, whatever $a$ we consider. To see this, you have to come back to the very definition : consider the *partial* integral $\int_a^X(-1)\,dx=a-X\underset{X\to+\infty}{\rightarrow}-\infty$ and also $\int_a^Xe^{-x}\,dx=e^{-a}-e^{-X}\underset{X\to+\infty}{\rightarrow}e^{-a}$ – 2017-02-16
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    what if it was $0 \leq f(x) \leq g(x)$ where the function is positive / continuous on all $ x \in [1,\infty)$, then if $\int_{1}^{\infty} g(x) dx$ converges wouldn't $\int_{1}^{\infty} f(x)dx$ converge as well. Or viceversa with diverge? (Isn't this the comparison test) – 2017-02-16
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    Yes, if you suppose $0\le f(x)\le g(x)$ for all $x\ge0$, now (b) becomes true ! – 2017-02-16
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    Okay I get it, thank you. One more question. So if it was $0 \leq f(x) \leq g(x)$ for b) this would still be true even for [0, $\infty$)? – 2017-02-16
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    If course and for any interval $a,+\infty$ – 2017-02-16