Evaluate the definite integral from 1 to -3 of (-x/2+1) dx

Question

Definite Integrals Practice

Answer 1

Given that the definite integral is integral from 1 to -3 of (-x/2+1).

To evaluate the given definite integral and find its value.

Consider the function f(x) = (-x/2+1).

Integrate the function f(x) with respect to x is as follows.

f(x)=(-x/2+1)

∫ f(x)dx = ∫ (-x/2+1) dx

= [-x⁽¹⁺¹⁾ / 2(1+1)+x]

= [-x² / 2*2)+x]

∫ f(x)dx = - x² / 4+x

Now apply the integral values from 1 to -3 in the above obtained value as shown below.∫

∫ f(x) dx = [(-1² / 4+1) -(-(-3)² / 4+(-3))]

= [(-1/4+1)-(-9/4-3)]

= [-1/4+1+9/4+3]

= [ 8/4 + 4]

= 2+4

∫ f(x)dx = 6

Thus the value of the given integral is 6.

Answer 2

f(x) = -x/2 + 1

F(x) = -x^2/4 +x

evaluate from -3 to 1

F(1) - F(-3) = -1/4 +1 - (-9/4 -3) =2+4= 6

check the answer

graph the line, it's a downward sloping line with y intercept =1 and x intercept = 2

use geometry to calculate the area under the line above the x axis between x=-3 and x=1

a quadrilateral with

base=1- -3 = 4, height = (5/2 + 1/2)/2 = 1.5 average.

4x1.5 = 6