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Which of the following improper integrals are convergent?

a.$\quad\displaystyle \int_{1}^{\infty} \frac{dx}{\sqrt{x^3+ 2x + 2}}$

b. $\quad\displaystyle \int_{0}^{5}\frac{dx}{(x^2− 5x + 6)}$

c.$\quad\displaystyle \int_{0}^{5} \frac{dx}{\sqrt[\large 3]{7x + 2x^4}}$

how can I able to solve this.thanks for your help

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1 Answer 1

The first one is convergent. Because $$\lim_{x\to\infty}x^{\frac{3}{2}}\cdot\frac{1}{\sqrt{x^3+2x+2}}=1$$ and we know when the resulted limit is finite when the power of $x$ is greater than $1$ (I mean that $x^{\frac{3}{2}}$ ), the improper integral $\int_a^{\infty}$ converges.

About the second. If we divide it to 3 parts, we have: $$\int_0^5=\int_0^2+\int_2^3+\int_3^5$$ Consider the first part: $$\int_0^2\frac{dx}{x^2-5x+6}=\int_0^2\frac{dx}{(x-2)(x-3)}$$ You see that if we consider $(x-2)^1$ and multiply it to the integrand $\frac{1}{(x-2)(x-3)}$ and take the limit of the result when $x\to 2^-$, then: $$\lim_{x\to 2^-}(x-2)^1\frac{1}{(x-2)(x-3)}=-1\neq0$$ whenever you can find a power like $1$ (I mean in $(x-2)^1$) and so find the above limit non zero, the improper integral diverges. So the second one is divergent.

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Helpful approach, covering lots of territory! +1 –  amWhy Feb 18 '13 at 0:12
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