# Solve The Negaive Definite Integral. Solve This When a=3, b=1 and f(x)=(x^2+x)

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### 3 Answers

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**

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! :).

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.