Question

Computing Indefinite Integrals:
- Evaluate $\displaystyle \int{{4{x^6} - 2{x^3} + 7x - 4\,dx}}$

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Answer 1

Answer:

$\displaystyle \int {4x^6 - 2x^3 + 7x - 4} \, dx = \boxed{ \frac{4x^7}{7} - \frac{x^4}{2} + \frac{7x^2}{2} - 4x + C }$

General Formulas and Concepts:
Calculus

Integration

• Integrals

Integration Rule [Reverse Power Rule]:
$\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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

Integration Property [Addition/Subtraction]:
$\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Step-by-step explanation:

Step 1: Define

Identify given integral.

$\displaystyle \int {4x^6 - 2x^3 + 7x - 4} \, dx$

Step 2: Integrate

We can find the indefinite integral using basic integration techniques listed under "Calculus":

1. [Integral] Rewrite [Integration Property - Addition/Subtraction]:
   $\displaystyle \begin{aligned}\int {4x^6 - 2x^3 + 7x - 4} \, dx & = \int {4x^6} \, dx - \int {2x^3} \, dx + \int {7x} \, dx - \int {4} \, dx \leftarrow \\\end{aligned}$
2. [Integrals] Rewrite [Integration Property - Multiplied Constant]:
   $\displaystyle \begin{aligned}\int {4x^6 - 2x^3 + 7x - 4} \, dx & = \int {4x^6} \, dx - \int {2x^3} \, dx + \int {7x} \, dx - \int {4} \, dx \\& = 4 \int {x^6} \, dx - 2\int {x^3} \, dx + 7 \int {x} \, dx - 4 \int {} \, dx \leftarrow \\\end{aligned}$
3. [Integrals] Apply Integration Rule [Reverse Power Rule]:
   $\displaystyle \begin{aligned}\int {4x^6 - 2x^3 + 7x - 4} \, dx & = \int {4x^6} \, dx - \int {2x^3} \, dx + \int {7x} \, dx - \int {4} \, dx \\& = 4 \int {x^6} \, dx - 2\int {x^3} \, dx + 7 \int {x} \, dx - 4 \int {} \, dx \\& = 4 \bigg( \frac{x^7}{7} \bigg) - 2 \bigg( \frac{x^4}{4} \bigg) + 7 \bigg( \frac{x^2}{2} \bigg) - 4x + C \leftarrow \\\end{aligned}$
4. Simplify:
   $\displaystyle \begin{aligned}\int {4x^6 - 2x^3 + 7x - 4} \, dx & = \int {4x^6} \, dx - \int {2x^3} \, dx + \int {7x} \, dx - \int {4} \, dx \\& = 4 \int {x^6} \, dx - 2\int {x^3} \, dx + 7 \int {x} \, dx - 4 \int {} \, dx \\& = 4 \bigg( \frac{x^7}{7} \bigg) - 2 \bigg( \frac{x^4}{4} \bigg) + 7 \bigg( \frac{x^2}{2} \bigg) - 4x + C \\& = \boxed{ \frac{4x^7}{7} - \frac{x^4}{2} + \frac{7x^2}{2} - 4x + C } \leftarrow \\\end{aligned}$

∴ we have found the indefinite integral.

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

Unit: Integration

Answer 2

Answer:

$\dfrac{4}{7}x^7-\dfrac{1}{2}x^4+\dfrac{7}{2}x^2-4x+\text{C}$

Step-by-step explanation:

Integration is the "opposite" of differentiation.

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration (C).

$\boxed{\begin{minipage}{5 cm}\underline{Integrating a constant}\\\\ \displaystyle \int n\:\text{d}x=nx+\text{C} \\(where n is any constant value)\end{minipage}}$

Just add an x to the constant.

$\boxed{\begin{minipage}{4 cm}\underline{Integrating x^n }\\\\ \displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C} \end{minipage}}$

Increase the power by 1, then divide by the new power.

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\ \displaystyle \int ax^n\:\text{d}x=a \int x^n \:\text{d}x \end{minipage}}$

If the terms are multiplied by constants, take them outside the integral.

Given indefinite integral:

$\displaystyle \int (4x^6-2x^3+7x-4)\:dx$

Integrate each term separately and take the constants outside the integrals:

$\implies \displaystyle 4 \int x^6\:\text{d}x - 2 \int x^3 \:\text{d}x + 7 \int x \:\text{d}x - \int 4 \:\text{d}x$

Integrate, using the rules given above:

$\implies 4 \cdot \dfrac{x^7}{7}-2 \cdot \dfrac{x^4}{4}+7 \cdot \dfrac{x^2}{2}-4x+\text{C}$

Simplify:

$\implies \dfrac{4}{7}x^7-\dfrac{1}{2}x^4+\dfrac{7}{2}x^2-4x+\text{C}$

As one calculation:

$\begin{aligned}\displaystyle \int (4x^6-2x^3+7x-4)\:dx & = \displaystyle 4 \int x^6\:\text{d}x - 2 \int x^3 \:\text{d}x + 7 \int x \:\text{d}x - \int 4 \:\text{d}x\\\\& = 4 \cdot \dfrac{x^7}{7}-2 \cdot \dfrac{x^4}{4}+7 \cdot \dfrac{x^2}{2}-4x+\text{C}\\\\& = \dfrac{4}{7}x^7-\dfrac{1}{2}x^4+\dfrac{7}{2}x^2-4x+\text{C}\end{aligned}$