Question

Find the integral, limit ( 0 to infinity)
∫[2e-x] dx
[] isintegral part function

Answer 1

Answer:

The given integral is diverging as it results infinity.

Step-by-step explanation:

Given function of the integral is:

$\displaystyle\int _0^{\infty}[2e-x]dx$

Taking $b\rightarrow \infty.$

Hence we may write the given function as:

$\lim_{b \to \infty} \displaystyle\int _0^{b}(2e-x)dx\\= \lim_{b \to \infty} \displaystyle\int _0^{b}2e\ dx- \lim_{b \to \infty} \displaystyle\int _0^{b}x\ dx\\= \lim_{b \to \infty} [2e\ x]_0^b- \lim_{b \to \infty} [\dfrac{x^2}{2}]_0^b\\= \lim_{b \to \infty} [2e\ b-0]- \lim_{b \to \infty} \dfrac{1}{2}[b^2-0]$

Substituting the value of $b\rightarrow \infty$ we get:

$=\infty$

Therefore the given function is diverging as it result a infinite value.

Answer 2

Answer:

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