Solve The Negaive Definite Integral. Solve This When a=3, b=1 and f(x)=(x^2+x)
3 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
To find the value of the integral of f(x)=(x^2+x), between the limits 1 and 3, we find the integral
Int[ f(x) dx ], x = 1 to x = 3
=> (x^3/3 + x^2/2), x = 1 to x = 3
=> 3^3/3 + 3^2/2 - 1^3/3 - 1^2/2
=> 9 + 9/2 - 1/3 - 1/2
=> 38/3
The required value if the definite integral is 38/3
iheartzyou | Student, Undergraduate | (Level 1) eNoter
I Did It A Different Way Before I Posted This Discussion And Got 38/3 As My Answer.
But Both Answers Are Still Correct I Assume.
Thanks Alot For You Help! :).
giorgiana1976 | College Teacher | (Level 3) Valedictorian
To solve the definite integral, we'll use the following convention: we'll reverse the limits of integration.
Int f(x)dx, from a to b = -Int f(x)dx, from b to a
In our case, we'll have:
Int (x^2+x)dx, from 3 to 1 = - Int (x^2+x)dx, from 1 to 3
Int (x^2+x)dx = Int x^2dx + Int xdx
Int (x^2+x)dx = x^3/3 + x^2/2
We'll apply Leibniz-Newton formula:
- Int (x^2+x)dx = -[F(3) - F(1)]
F(1) = 1/3 + 1/2 = 5/6
F(3) = 27/3 + 9/2
F(3) = 9 + 9/2
F(3) = 27/2
F(1) - F(3) = 5/6 - 27/2 = -(5-81)/6 = -76/6
The definite integral is 76/6.