I"ve been trying to incorporate that function, but it looks choose I"m absent something, so can any type of one please present me exactly how to integrate $e^\sqrtx$, in order to correct my procedure.Thanks  You"re going to want to begin with a $u$-substitution $u=\sqrtx$. Notification then that $u^2=x$, for this reason $2udu=dx$. Then$$\int e^\sqrtxdx=2\int e^u u\ du.$$

You deserve to then continue by integration by parts.

To gain started, friend can collection $w=u$ and also $dv=e^u\ du$, and also use the fact that $\int w\ dv= wv-\int v\ dw$. Lastly, don"t forget come substitute ago to get the integral in regards to $x$.

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Continuing, $dw=du$ and $v=e^u$. Then $$\int w\ dv= wv-\int v\ dw = e^u u-\int e^u\ du=e^u u-e^u.$$So$$\int e^\sqrtxdx=2\int e^u u\ du=2e^uu-2e^u+C=2e^\sqrtx(\sqrtx-1)+C$$after earlier substituting $u=\sqrtx$ and including a possible consistent term.

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answered Sep 26 "11 at 21:53 yunoneyunone
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You can integrate $e^\sqrtx$ by substituting any solitary variable in ar of $\sqrtx$.

Since, $e^\sqrtx$ does no look choose $e^x$ due to the visibility of square root, therefore it have the right to not be incorporated by common methods.

$\int e^\sqrtx \,dx\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...(1)$

To incorporate $e^\sqrtx$, you must substitute any type of variable, to speak $t$ in location of $\sqrtx.$

Now, $e^\sqrtx$ becomes $e^t$ which is in the form of $e^x$, so it can be integrated easily.

But, when you instead of $t=$$\sqrtx$, then equation $(1)$ becomes :

$\int e^t \,dx$

and $t$ have the right to not be incorporated with respect to $x.$

So, change $dx$ by examining its value as –

$t=\sqrtx$

$t^2=x$

On differentiation

$2tdt=dx$

Thus, $\int e^\sqrtx\,dx=\int e^t \cdot 2t\,dt$

$=2\int e^t \cdot t\,dt \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; ...(2)$

Now, you have to combine a product of two functions, so you require the formula because that integration through parts.

$\int u\,dv=uv - \int v\,du$

Here, $u$ and $v$ must be assumed by adhering to the order of ILATE (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric and also Exponential) from left come right.

For example : If you have a product of 2 terms, to speak $x \cdot sec^2 x \,dx,$ then according to the order of ILATE, algebraic hatchet $(x)$ comes before the trigonometric hatchet $(sec^2 x).$

Therefore, friend should think about the algebraic ax $(x)$ as $u$ and also the trigonometric ax $(sec^2 x\,dx)$ together $dv$.

Now, going back to the original product i.e., $e^t \cdot t\,dt$,we uncover that $t$ is an algebraic hatchet which comes before the exponential term, $e^t.$

So, allow $u=t$ and $dv= e^t\,dt$

Now, you require the values of $v$ and $du$ because that the formula.

$\int u\,dv=uv–\int v \,du$

Therefore, distinguish $u=t$ and also integrate $dv=e^t\,dt$

$du=dt$ and $v=e^t$

Put all these values in the formula –

$\int e^t \cdot t \,dt=t \cdot e^t – \int e^t\,dt$

$=t \cdot e^t – e^t$

Now, from equation (2)

$\int e^\sqrtx\,dx = 2 \int e^t \cdot t \,dt$

$=2 (t \cdot e^t – e^t )+c$

Take $e^t$ the end of the parenthesis

$=2 e^t (t–1)+c$

Put the worth of $t$ to gain the price in regards to $x.$

So, $\int e^\sqrtx\,dx = 2 e^\sqrtx (\sqrtx–1)+c$

Here, $c$ is the constant.

This, is the integral of $\int e^\sqrtx \,dx.$ You can likewise integrate different types of functions using substitution method, i beg your pardon otherwise can’t be integrated by common methods.

Let’s integrate $cos^3 x$ to understand the substitution dominion better.

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$\int cos^3 x\,dx$

$cos^3 x$ can not be combined directly. So, we must factorize it together :

$\int cos^2 x \cdot cos x\,dx$

Now, if us substitute $u=cos x$

then on differentiation, we get :

$du=–sin x\,dx$

But, we don’t have $–sin x$ in both the the factors .

i.e., $cos^2 x, cos x$

So, us can’t substitute $u=cos x$

For substitution, us should have actually both the functions, $sin x$ and $cos x$

$\because\, sin^2 x + cos^2 x = 1$

Subtracting $sin^2 x$ from both the sides

$\Rightarrow sin^2 x + cos^2 x – sin^2 x = 1 – sin^2 x$

$\Rightarrow sin^2 x – sin^2 x + cos^2 x = 1 – sin^2 x$

$cos^2x = 1 – sin^2 x$

So, we deserve to replace $cos^2 x$ by $1 – sin^2 x$

Then $\int cos^2 x \cdot cos x\,dx$ becomes, $\int (1–sin^2 x) \cdot cos x\, dx$