New Notes Blog: `int x^3e^x dx` Find the indefinite integral

Sunday, 27 September 2015

$\displaystyle\int{{x}}^{{3}}{{e}}^{{x}}{\left.{d}{x}\right.}$ Find the indefinite integral

To evaluate the integral: $\displaystyle\int{{x}}^{{3}}{{e}}^{{x}}{\left.{d}{x}\right.}$ , we may apply "integration by parts": $\displaystyle\int{u}\cdot{d}{v}={u}{v}-\int{v}{d}{u}$ .

Let: $\displaystyle{u}={{x}}^{{3}}$ then $\displaystyle{d}{u}={3}{{x}}^{{2}}{\left.{d}{x}\right.}$

$\displaystyle{d}{v}={{e}}^{{x}}{\left.{d}{x}\right.}$ then $\displaystyle{v}=\int{{e}}^{{x}}{\left.{d}{x}\right.}={{e}}^{{x}}$ .

Apply the formula for integration by parts, we get:

$\displaystyle\int{{x}}^{{3}}{{e}}^{{x}}{\left.{d}{x}\right.}={{x}}^{{3}}{{e}}^{{x}}-\int{3}{{x}}^{{2}}{{e}}^{{x}}{\left.{d}{x}\right.}$ .

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To evaluate the integral: $\displaystyle\int{{x}}^{{3}}{{e}}^{{x}}{\left.{d}{x}\right.}$ , we may apply "integration by parts": $\displaystyle\int{u}\cdot{d}{v}={u}{v}-\int{v}{d}{u}$ .

Let: $\displaystyle{u}={{x}}^{{3}}$ then $\displaystyle{d}{u}={3}{{x}}^{{2}}{\left.{d}{x}\right.}$

$\displaystyle{d}{v}={{e}}^{{x}}{\left.{d}{x}\right.}$ then $\displaystyle{v}=\int{{e}}^{{x}}{\left.{d}{x}\right.}={{e}}^{{x}}$ .

Apply the formula for integration by parts, we get:

$\displaystyle\int{{x}}^{{3}}{{e}}^{{x}}{\left.{d}{x}\right.}={{x}}^{{3}}{{e}}^{{x}}-\int{3}{{x}}^{{2}}{{e}}^{{x}}{\left.{d}{x}\right.}$ .

$\displaystyle={{x}}^{{3}}{{e}}^{{x}}-{3}\int{{x}}^{{2}}{{e}}^{{x}}{\left.{d}{x}\right.}.$

To evaluate $\displaystyle\int{{x}}^{{2}}{{e}}^{{x}}{\left.{d}{x}\right.}$ , we apply another set of integration by parts.

Let: $\displaystyle{u}={{x}}^{{2}}$ then $\displaystyle{d}{u}={2}{x}{\left.{d}{x}\right.}$

$\displaystyle{v}={{e}}^{{x}}{\left.{d}{x}\right.}$ then $\displaystyle{d}{v}={{e}}^{{x}}$

The integral becomes:

$\displaystyle\int{{x}}^{{2}}{{e}}^{{x}}{\left.{d}{x}\right.}={{x}}^{{2}}{{e}}^{{x}}-\int{2}{x}{{e}}^{{x}}{\left.{d}{x}\right.}$

Another set of integration by parts by letting:

$\displaystyle{u}={2}{x}$ then $\displaystyle{d}{u}={2}{\left.{d}{x}\right.}$

$\displaystyle{v}={{e}}^{{x}}{\left.{d}{x}\right.}$ then $\displaystyle{d}{v}={{e}}^{{x}}$

$\displaystyle\int{2}{x}{{e}}^{{x}}{\left.{d}{x}\right.}={2}{x}{{e}}^{{x}}-\int{2}{{e}}^{{x}}{\left.{d}{x}\right.}$

$\displaystyle={2}{x}{{e}}^{{x}}-{2}{{e}}^{{x}}+{C}$

Using $\displaystyle\int{2}{x}{{e}}^{{x}}{\left.{d}{x}\right.}={2}{x}{{e}}^{{x}}-{2}{{e}}^{{x}}+{C}$ , we get:

$\displaystyle\int{{x}}^{{2}}{{e}}^{{x}}{\left.{d}{x}\right.}={{x}}^{{2}}{{e}}^{{x}}-\int{2}{x}{{e}}^{{x}}{\left.{d}{x}\right.}$

$\displaystyle={{x}}^{{2}}{{e}}^{{x}}-{\left[{2}{x}{{e}}^{{x}}-{2}{{e}}^{{x}}\right]}+{C}$

$\displaystyle={{x}}^{{2}}{{e}}^{{x}}-{2}{x}{{e}}^{{x}}+{2}{{e}}^{{x}}+{C}$

Then,

$\displaystyle\int{{x}}^{{3}}{{e}}^{{x}}{\left.{d}{x}\right.}={{x}}^{{3}}{{e}}^{{x}}-{3}\int{{x}}^{{2}}{{e}}^{{x}}{\left.{d}{x}\right.}$ .

$\displaystyle={{x}}^{{3}}{{e}}^{{x}}-{3}{\left[{{x}}^{{2}}{{e}}^{{x}}-{2}{x}{{e}}^{{x}}+{2}{{e}}^{{x}}\right]}+{C}$

$\displaystyle={{x}}^{{3}}{{e}}^{{x}}-{3}{{x}}^{{2}}{{e}}^{{x}}+{6}{x}{{e}}^{{x}}-{6}{{e}}^{{x}}+{C}$

New Notes Blog

Sunday, 27 September 2015

$\displaystyle\int{{x}}^{{3}}{{e}}^{{x}}{\left.{d}{x}\right.}$ Find the indefinite integral

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Sunday, 27 September 2015

Find the indefinite integral

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In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle\int{{x}}^{{3}}{{e}}^{{x}}{\left.{d}{x}\right.}$ Find the indefinite integral

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?