Answer:
[tex]\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx = 32[/tex]
General Formulas and Concepts:
Calculus
Integration
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
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
