given �ç 1to-1 f(x)dx=0 and �ç 1to0 f(x)dx=5, compute the following
a.�ç 0to-1 f(x)dx
b.�ç 0to1 3f(x)dx=0
c.�ç 1to-1 (f(x)+7)dx=0
ç is the integral sign
Using I for integral,
I f(x) dx [1,-1] = I f(x) dx [1,0] + I f(x) dx [0,-1]
I f(x) dx [0,1] = -I f(x) dx [1,0]
I 3f(x) dx = 3 * I f(x) dx
I (f(x)+7) dx = I f(x) dx + I 7 dx
Now just plug and chug...
To find the solutions to these definite integrals, we can use the properties of definite integrals and the given information.
a. ∫[0 to -1] f(x) dx:
Since we know that ∫[1 to -1] f(x) dx = 0, we can break down the integral into two parts:
∫[1 to 0] f(x) dx + ∫[0 to -1] f(x) dx = 0
We are given that ∫[1 to 0] f(x) dx = 5, so we can substitute this value into the equation:
5 + ∫[0 to -1] f(x) dx = 0
Subtracting 5 from both sides:
∫[0 to -1] f(x) dx = -5
b. ∫[0 to 1] 3f(x) dx:
To find this integral, we can use the linearity property of definite integrals, which states that the integral of the sum of two functions is equal to the sum of their integrals.
∫[0 to 1] 3f(x) dx = 3 * ∫[0 to 1] f(x) dx
We are given that ∫[1 to 0] f(x) dx = 5, so we can substitute this value:
3 * 5 = 15
Thus, ∫[0 to 1] 3f(x) dx is equal to 15.
c. ∫[1 to -1] (f(x) + 7) dx:
Using the linearity property of definite integrals, we can rewrite the integral:
∫[1 to -1] f(x) dx + ∫[1 to -1] 7 dx = 0
We know that ∫[1 to -1] f(x) dx = 0 from the given information. We can simplify the equation:
0 + 7 * ∫[1 to -1] dx = 0
The integral of a constant over any interval is just the product of the constant and the width of the interval. Since we are integrating from 1 to -1, the width is 1 - (-1) = 2:
7 * 2 = 14
Thus, ∫[1 to -1] (f(x) + 7) dx is equal to 14.