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Yet another Calculus question...

This time, it's about integrals/antiderivatives. The question asks to find the value of [tex]\int\limits^6_0 {[g(x)+2]} \, dx[/tex]. Since the derivative of 2 is 0, can I directly find [tex]\int\limits^6_0 {g(x)} \, dx[/tex]? Thanks in advance!

Sagot :

Space

Answer:

[tex]\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx = 32[/tex]

General Formulas and Concepts:

Calculus

Integration

  • Integrals

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

Integration Rule [Fundamental Theorem of Calculus 1]:                                     [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

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

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

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle \int\limits^6_0 {g(x)} \, dx = 20[/tex]

[tex]\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx[/tex]

Step 2: Integrate

  1. [Integral] Rewrite [Integration Property - Addition/Subtraction]:               [tex]\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx = \int\limits^6_0 {g(x)} \, dx + \int\limits^6_0 {2} \, dx[/tex]
  2. [2nd Integral] Rewrite [Integration Property - Multiplied Constant]:         [tex]\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx = \int\limits^6_0 {g(x)} \, dx + 2\int\limits^6_0 {} \, dx[/tex]
  3. [1st Integral] Substitute in value:                                                                   [tex]\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx = 20 + 2\int\limits^6_0 {} \, dx[/tex]
  4. [Integral] Reverse Power Rule:                                                                     [tex]\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx = 20 + 2(x) \bigg| \limits^6_0[/tex]
  5. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           [tex]\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx = 20 + 2(6)[/tex]
  6. Simplify:                                                                                                         [tex]\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx = 32[/tex]

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

Unit: Integration

Answer:

32

(I think I remember your other information correctly.)

Step-by-step explanation:

I think you said you were given

*Integral from x=-5 to x=0 of g was -14

*Integral from x=-5 to x=6 of g was 6

Asked to find integral( g(x) + 2 , from x=0 to x=6)

Yes this can be split into two integrals:

Integral(g(x), x=0 to x=6) + Integral(2, x=0 to x=6)

The last integral is easier... the antiderivative or 2 is 2x. So evaluate 2x as the limits and subtract. Always plug in the top limit first. 2(6)-2(0)=12-0=12

Let's start with the bigger interval from x=-5 to x=6 which was 6... and since we want to get rid of the interval from x=-5 to x=0 to find the integral of g from x=0 to 6, all we must do is do 6-(-14)=20.

Integral( g(x) + 2 , from x=0 to x=6)

=

Integral(g(x), x=0 to x=6) + Integral(2, x=0 to x=6)

=

20+12

=

32

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