Question

Evaluate the integral of dx/(x + 2)from -6 to -10.​

Answer 1

EXPLANATION.

$\sf \implies\int\limits_{-6}^{-10}\dfrac{dx}{(x + 2)}.$

By using the substitutions method, we get.

Substitute (x + 2) = t.

Differentiate w.r.t x, we get.

⇒ dx = dt.

Put the values in equation, we get.

$\sf \implies \int\limits_{-6}^{-10}\dfrac{dt}{t}$

As we know that,

⇒ ∫dt/t = ln | t | + c.

$\sf \implies \int\limits_{-6}^{-10} ln | t| + c.$

Put the value of t in equation, we get.

$\sf \implies [ln |x + 2| ]\limits_{-6}^{-10}$

⇒ [ln | -10 + 2|] -[ ln | - 6 + 2 |].

⇒ [ ln(-8) ] - [ ln(-4) ].

⇒ ln(-8) + ln(4).

                                                                                       

MORE INFORMATION.

Definition [NEWTON-LEIBNITZ FORM].

Let f'' is the function of x defined on [a, b] and d[f(x)]/dx = Ф(x) and a and b, are two values independent of variable x, then for all values of x  in domain of f'' then,

$\sf\implies\int\limits_{a}^{b}\phi(x)dx = [f(x)]\limits_{a}^{b} = f(b) - f(a).$

Answer 2

${ \boxed {\bf{Given \: Question}}}$

Evaluate the integral of dx/(x + 2) from -6 to -10.

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${ \boxed {\bf{Formula \: used :- }}}$

$\bf\int\ \dfrac{1}{x}dx = logx \: + c$

$\bf \: log(x) - log(y) = log( \dfrac{x}{y} )$

$\ \blue{ \large \bf \: Solution :- }$

$\bf\int\limits_{ - 6}^{ - 10} \dfrac{1}{x + 2}dx$

$\bf\implies \:[log |x + 2| ]_{-6}^{-10}$

$\bf\implies \: log | - 10 + 2| - log | - 6 + 2|$

$\bf\implies \: log | -8| - log | - 4|$

$\bf\implies \: log | 8| - log | 4|$

$\bf\implies \:log\dfrac{8}{4} = log2$

${ \boxed {\bf\implies \:{\bf\int\limits_{ - 6}^{ - 10} \dfrac{1}{x + 2}dx = log2}}}$

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