Integral from 1 to 3 of ( 19 + 2x - 20x ) dx

thanks

EDIT:

that 20x is really 20x^3

separate it into three integrals

19 dx --> 19 x ---> )19*3 - 19*1
2x^3 dx --> (1/2)x^4--> .5(27-1)
-20xdx--> -10x^2-->10(1-9)

Ok, first you integrate the problem so it equals 19x + x^2 - 5x^4

Next I'll write this out then show you. You then plug in x = 3 to this equation and then plug in x = 1

Subtract the equation of x = 1 from the equation x = 3.

(19(3) - (3)^2 - 5(3)^4) - (19(1) - (1)^2 - 5(1)^4) = -354

2x^3 dx --> (1/2)x^4--> .5(27-1)

should be
20x^3 dx --> (10/2)x^4--> 5(81-1)

To compute the integral of the given function, we can use the power rule and the constant rule.

Step 1: Distribute the integral over the terms in the integrand:
∫ (19 + 2x - 20x) dx

Step 2: Split the integral into separate integrals for each term:
∫ 19 dx + ∫ 2x dx - ∫ 20x dx

Step 3: Apply the power rule and constant rule to solve each integral.

For the first term, ∫ 19 dx, the integral of a constant is the constant multiplied by the variable of integration:
∫ 19 dx = 19x + C (where C is the constant of integration)

For the second term, ∫ 2x dx, we use the power rule for integrals:
∫ 2x dx = x^2 + C

For the third term, ∫ 20x dx, we apply both the power rule and the constant rule:
∫ 20x dx = 10x^2 + C

Step 4: Combine the results from each integral:
∫ (19 + 2x - 20x) dx = ∫ 19 dx + ∫ 2x dx - ∫ 20x dx
= 19x + x^2 - 10x^2 + C
= -9x^2 + 19x + C

Now, to evaluate the definite integral from 1 to 3, substitute the upper and lower limits of integration:
∫ (19 + 2x - 20x) dx evaluated from 1 to 3
= (-9(3)^2 + 19(3) + C) - (-9(1)^2 + 19(1) + C)
= (-9(9) + 19(3) + C) - (-9(1) + 19(1) + C)
= (-81 + 57 + C) - (-9 + 19 + C)
= -24 - (-9)
= -24 + 9
= -15

Therefore, the value of the integral from 1 to 3 of (19 + 2x - 20x) dx is -15.