Question

Yet another Calculus question...

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

Answer 1

32

(I think I remember your other information correctly.)

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

Answer 2

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

General Formulas and Concepts:

Calculus

Integration

• Integrals

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

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

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`

Explanation:

Step 1: Define

Identify

`\displaystyle \int\limits^6_0 {g(x)} \, dx = 20`

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

Step 2: Integrate

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

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

Unit: Integration