How do you evaluate the definite integral #int (x-sqrtx)/3dx# from [0,1]?

Question

Emma Aldridge · Accepted Answer

#-1/18.# By the Fundamental Theorem of Calculus, #int_0^1 (x-sqrtx)/3 dx=1/3[x^2/2-x^(1/2+1)/(1/2+1)]_0^1# #=1/3[x^2/2-(2/3)x^(3/2)]# #=1/3{{1/2-2/3}-0]# #=1/3(-1/6)# #=-1/18.# Enjoy Maths.

Cameron Alexander · Answer

undefined To evaluate the definite integral ∫ (x - √x)/3 dx from 0 to 1, you can follow these steps:

1. Split the integral into two separate integrals:
   ∫ (x - √x)/3 dx = ∫ (x/3 - √x/3) dx

2. Integrate each term separately:
   ∫ x/3 dx = (1/3) * ∫ x dx = (1/3) * (x^2/2) = x^2/6
   ∫ √x/3 dx = (1/3) * ∫ √x dx = (1/3) * (2/3) * x^(3/2) = (2/9) * x^(3/2)

3. Evaluate each integral from 0 to 1:
   Substitute the upper limit (1) and subtract the result of substituting the lower limit (0) for each integral.

4. Combine the results:
   ∫ (x - √x)/3 dx = [x^2/6 - (2/9) * x^(3/2)] evaluated from 0 to 1.

5. Substitute the upper limit:
   (1^2/6 - (2/9) * 1^(3/2)) = (1/6 - 2/9)

6. Substitute the lower limit:
   (0^2/6 - (2/9) * 0^(3/2)) = (0 - 0) = 0

7. Subtract the results:
   (1/6 - 2/9) - 0 = (1/6 - 2/9)

8. Find a common denominator and subtract the fractions:
   (3/18 - 4/18) = -1/18

Therefore, the value of the definite integral ∫ (x - √x)/3 dx from 0 to 1 is -1/18.