calculus single variable

Question

The area between the x -axis and the graph of f(x)=x^2-5x+6 = on [1,4] is

Answer 1

Set up your definite integral with a lower limit of 1 and upper limit of 4. Solve the integral through anti differentiation and the Fundamental Theorem of Calculus.

∫(x²-5x+6)dx = x³/3-5x²/2+6x [1,4] = [4³/3-5*4²/2+6*4] - [1³/3-5*1²/2+6*1] = 16/3 - 23/6 = 3/2

Now, there could be a different answer if you want to count the areas between this curve and the x-axis as POSITIVE. There are 3 regions to consider: [1,2] = positive, [2,3] = negative because it's below the x-axis, and [3,4]= positive. So if you want to count them ALL as positive areas, you must solve 3 separate integrals over these intervals, and the middle one from [2,3] will be the ABSOLUTE VALUE or opposite of the function. Here's the set up:

∫(x²-5x+6)dx [1,2] + ∫-(x²-5x+6)dx [2,3] + ∫(x²-5x+6)dx [3,4] = 5/6 + 1/6 + 5/6

Solve each of these by the Fundamental Theorem of Calculus. Final answer: 11/6

Answer 2

The area under the curve is found by taking the integral F(x) = ∫(x²-5x+6)dx and then computing F(4)-F(1).

F(x) = (1/3)x³-(5/2)x²+6x

Compute F(4)-F(1): ((1/3)*4³ -(5/2)*4² +6*4 ) - ((1/3)*1³ -(5/2)*1² +6*1 )

This simplifies to 1.5