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Chapter 5: Q 36. (page 429)

Solve the integral: $鈭� \frac{x^{2}}{e^{x}} d x$

Short Answer

Expert verified

The required answer is $(- \frac{x}{e^{x}}) - \frac{x}{2 e^{x}} - \frac{1}{2} e^{- x} + c$ .

Step by step solution

01

Step 1. Given information.

We have given integral is $鈭� \frac{x^{2}}{e^{x}} d x$ .

02

Step 2. Solve the integration by parts .

We have, $u = x^{2}, d v = \frac{d x}{e^{x}}$

$u = x^{2} d u = 2 x d x \frac{d u}{2} = x d x$

and

$d v = \frac{d x}{e^{x}} v = 鈭� \frac{d x}{e^{x}} v = - \frac{1}{e^{x}}$

The formula of integration by parts is $鈭� u d v = u v - 鈭� v d u$ .

role="math" localid="1648713324601" $鈭� \frac{x^{2}}{e^{x}} d x = [x^{2} (- \frac{1}{e^{x}}) - \frac{1}{2} 鈭� x (- \frac{1}{e^{x}}) d x] = (- \frac{x}{e^{x}}) + \frac{1}{2} 鈭� \frac{x}{e^{x}} d x = (- \frac{x}{e^{x}}) + \frac{1}{2} [x (- \frac{1}{e^{x}}) - 鈭� (- \frac{1}{e^{x}}) d x] = (- \frac{x}{e^{x}}) + \frac{1}{2} [(- \frac{x}{e^{x}}) + 鈭� e^{- x} d x]$

03

Step 3. Integration by parts.

We have, $u = - x d u = - d x$

$(- \frac{x}{e^{x}}) + \frac{1}{2} [(- \frac{x}{e^{x}}) + 鈭� e^{- x} d x] = (- \frac{x}{e^{x}}) + \frac{1}{2} [(- \frac{x}{e^{x}}) - e^{- x}] + c = (- \frac{x}{e^{x}}) - \frac{x}{2 e^{x}} - \frac{1}{2} e^{- x} + c$

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Most popular questions from this chapter

Solve the integral $鈭� x^{3} \sqrt{x^{2} - 1} d x$ three ways:

(a) with the substitution $u = x^{2} - 1,$ followed by back substitution;

(b) with integration by parts, choosing localid="1648814744993" $u = x^{2} and d v = x \sqrt{x^{2} - 1} d x;$

(c) with the trigonometric substitution x = sec u.

Explain why it makes sense to try the trigonometric substitution $x = \sec 鈦� u$ if an integrand involves the expression $\sqrt{x^{2} 鈭� 1}$

For each integral in Exercises 5鈥�8, write down three integrals that will have that form after a substitution of variables.

$鈭� \sin u d u$

Solve the integral: $鈭� {(x - e^{x})}^{2} d x$

Show by differentiating (and then using algebra) that $鈭� \cot 鈦� (\sin^{鈭� 1} 鈦� x)$ and $鈭� \frac{\sqrt{1 鈭� x^{2}}}{x}$ are both antiderivatives of $\frac{1}{x^{2} \sqrt{1 鈭� x^{2}}}$ . How can these two very different-looking functions be an antiderivative of the same function?

See all solutions

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