Question

Integration by substitution Using the substitution u=x'2-1, what is the value of the integral of 2xsqrt(x'2-1)dx from x=1 to x=2?

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\int\limits_1^2\:\left(\:2x\:\sqrt{x^2\:-\:1\:}\:\right)\:dx\:=\:3.464\:}}}$

Step-by-step-explanation:

The given function is

$\displaystyle{\sf\:2\:x\:\sqrt{x^2\:-\:1\:}\:}$

We have to integrate this function from x = 1 to x = 2.

Let us substitute,

$\displaystyle{\sf\:u\:=\:x^2\:-\:1}$

Differentiating both sides w.r.t. x, we get,

$\displaystyle{\sf\:\dfrac{d}{dx}\:(\:u\:)\:=\:\dfrac{d}{dx}\:(\:x^2\:-\:1\:)}$

$\displaystyle{\implies\sf\:\dfrac{du}{dx}\:=\:\dfrac{d}{dx}\:(\:x^2\:)\:-\:\dfrac{d}{dx}\:(\:1\:)}$

$\displaystyle{\implies\sf\:\dfrac{du}{dx}\:=\:2x\:-\:0}$

$\displaystyle{\implies\sf\:\dfrac{du}{dx}\:=\:2x}$

$\displaystyle{\implies\sf\:\dfrac{du}{2x}\:=\:dx}$

$\displaystyle{\implies\:\boxed{\blue{\sf\:dx\:=\:\dfrac{1}{2x}\:du\:}}}$

Now, lower limit,

When x = 1

u = x² - 1

⇒ u = 1² - 1

⇒ u = 1 - 1

⇒ u = 0

And, upper limit,

When x = 2,

u = x² - 1

⇒ u = 2² - 1

⇒ u = 4 - 1

⇒ u = 3

Now, let

$\displaystyle{\sf\:I\:=\:\int\limits_1^2\:\left(\:2x\:\sqrt{x^2\:-\:1\:}\:\right)\:dx}$

After substituting u = x² - 1, we get,

$\displaystyle{\sf\:I\:=\:\int\limits_0^3\:\left(\:\cancel{2x}\:\sqrt{u}\:\right)\:\dfrac{1}{\cancel{2x}}\:du}$

$\displaystyle{\implies\sf\:I\:=\:\int\limits_0^3\:\left(\:\sqrt{u}\:\right)\:du}$

$\displaystyle{\implies\sf\:I\:=\:\int\limits_0^3\:\left(\:u^{\frac{1}{2}}\:\right)\:du}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:\int\:x^n\:=\:\dfrac{x^{n\:+\:1}}{n\:+\:1}\:}}}$

$\displaystyle{\implies\sf\:I\:=\:\left[\:\dfrac{u^{\frac{1}{2}\:+\:1}}{\dfrac{1}{2}\:+\:1}\:\right]_0^3}$

$\displaystyle{\implies\sf\:I\:=\:\left[\:\dfrac{u^{\frac{1\:+\:2}{2}}}{\dfrac{1\:+\:2}{2}}\:\right]_0^3}$

$\displaystyle{\implies\sf\:I\:=\:\left[\:\dfrac{u^{\frac{3}{2}}}{\dfrac{3}{2}}\:\right]_0^3}$

$\displaystyle{\implies\sf\:I\:=\:\left[\:\dfrac{2u^{\frac{3}{2}}}{3}\:\right]_0^3}$

$\displaystyle{\implies\sf\:I\:=\:\dfrac{2\:\times\:3^{\frac{3}{2}}}{3}\:-\:\dfrac{2\:\times\:0^{\frac{3}{2}}}{3}\:}$

$\displaystyle{\implies\sf\:I\:=\:\dfrac{2\:\times\:(\:3^{\frac{1}{2}\:)^3}}{3}\:-\:\dfrac{2\:\times\:0}{3}}$

$\displaystyle{\implies\sf\:I\:=\:\dfrac{2\:\times\:(\:\sqrt{3}\:)^3}{3}\:-\:\dfrac{0}{3}}$

$\displaystyle{\implies\sf\:I\:=\:\dfrac{2\:\times\:\cancel{5.196}}{\cancel{3}}\:-\:0}$

$\displaystyle{\implies\sf\:I\:=\:2\:\times\:1.732}$

$\displaystyle{\implies\sf\:I\:=\:3.464}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\int\limits_1^2\:\left(\:2x\:\sqrt{x^2\:-\:1\:}\:\right)\:dx\:=\:3.464\:}}}}$