Mathematics, 14.01.2020 21:31 ggujjnh

Exercise is mixed —some require integration by parts, while others can be integrated by using techniques discussed in the chapter on integration.
∫x^3ex^4 dx.

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Answer from: freeman36

$\displaystyle \int {x^3e^{x^4}} \, dx = \frac{e^{x^4}}{4} + C$

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

Integrals[Indefinite Integrals] Integration Constant C

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int {x^3e^{x^4}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = x^4$ [u] Basic Power Rule: $\displaystyle du = 4x^3 \ dx$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {x^3e^{x^4}} \, dx = \frac{1}{4} \int {4x^3e^{x^4}} \, dx$ [Integral] U-Substitution: $\displaystyle \int {x^3e^{x^4}} \, dx = \frac{1}{4} \int {e^{u}} \, dx$ [Integral] Exponential Integration: $\displaystyle \int {x^3e^{x^4}} \, dx = \frac{e^{u}}{4} + C$ [u] Back-Substitute: $\displaystyle \int {x^3e^{x^4}} \, dx = \frac{e^{x^4}}{4} + C$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Answer from: Quest

$\begin {array}{c|c}\underline {\quad x\quad}& \underline{\quad y\quad}\\ -2& 4\\-1& 2\\0& 0\\1& 2\\ 2& 4\\\end{array}$

step-by-step explanation:

y = | -2(-2) | → y = | 4 | → y = 4 ⇒ (-2, 4)

y = | -2(-1) | → y = | 2 | → y = 2 ⇒ (-1, 2)

y = | -2(0) | → y = | 0 | → y = 0 ⇒ (0, 0)

y = | -2(1) | c → y = | -2 | → y = 2 ⇒ (1, 2)

y = | -2(2) | → y = | - 4 | → y = 4 ⇒ (2, 4)

Answer from: Quest

x= 1

step-by-step explanation: convert 5\frac{11}{12}5

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{5\times 12+11}{12}3.25+0.5x+1

5×12+11

2 simplify 5\times 125×12 to 6060

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{60+11}{12}3.25+0.5x+1

60+11

3 simplify 60+1160+11 to 7171

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{71}{12}3.25+0.5x+1

4 simplify \frac{5}{6}x

x to \frac{5x}{6}

3.25+0.5x+1\frac{1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+1

5 convert 1\frac{1}{3}1

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+\frac{1\times 3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

1×3+1

6 simplify 1\times 31×3 to 33

3.25+0.5x+\frac{3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

3+1

7 simplify 3+13+1 to 44

3.25+0.5x+\frac{4}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

8 simplify 3.25+0.5x+\frac{4}{3}+\frac{5x}{6}3.25+0.5x+

to \frac{13.75}{3}+\frac{4x}{3}

13.75

\frac{13.75}{3}+\frac{4x}{3}=\frac{71}{12}

13.75

9 simplify \frac{13.75}{3}

13.75

to 4.5833334.583333

4.583333+\frac{4x}{3}=\frac{71}{12}4.583333+

10 subtract 4.5833334.583333 from both sides

\frac{4x}{3}=\frac{71}{12}-4.583333

−4.583333

11 simplify \frac{71}{12}-4.583333

−4.583333 to \frac{4}{3}

\frac{4x}{3}=\frac{4}{3}

12 multiply both sides by 33

4x=\frac{4}{3}\times 34x=

×3

13 simplify \frac{4}{3}\times 3

×3 to \frac{12}{3}

4x=\frac{12}{3}4x=

14 simplify \frac{12}{3}

to 44

4x=44x=4

15 divide both sides by 44

x=1

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