Find the area between the parabola [tex]y=x^2-3[/tex] and the line [tex]y=x[/tex].
Answer (1)
Find the intersection points of y = x 2 − 3 and y = x by solving x 2 − x − 3 = 0 , resulting in x 1 ≈ − 1.303 and x 2 ≈ 2.303 .
Set up the definite integral to calculate the area between the curves: ∫ x 1 x 2 ( x − ( x 2 − 3 )) d x .
Evaluate the integral: ∫ − 1.303 2.303 ( x − x 2 + 3 ) d x = [ 2 x 2 − 3 x 3 + 3 x ] − 1.303 2.303 .
Calculate the area, which is approximately 7.812 .
Explanation
Problem Setup We are asked to find the area between the parabola y = x 2 − 3 and the line y = x . To do this, we need to find the points of intersection between the two curves, and then integrate the difference between the two functions over that interval.
Finding Intersection Points First, we need to find the intersection points of the parabola and the line. We set the two equations equal to each other: x 2 − 3 = x . Rearranging this gives us a quadratic equation: x 2 − x − 3 = 0 .
Solving for Intersection Points We can solve this quadratic equation using the quadratic formula: x = 2 a − b ± b 2 − 4 a c . In this case, a = 1 , b = − 1 , and c = − 3 . Plugging these values into the quadratic formula, we get: x = 2 ( 1 ) 1 ± ( − 1 ) 2 − 4 ( 1 ) ( − 3 ) = 2 1 ± 1 + 12 = 2 1 ± 13 . So the two intersection points are x 1 = 2 1 − 13 ≈ − 1.303 and x 2 = 2 1 + 13 ≈ 2.303 .
Setting up the Integral Now we need to find the area between the two curves. Since the line is above the parabola in the region of interest, we integrate the difference between the line and the parabola from x 1 to x 2 : A re a = ∫ x 1 x 2 ( x − ( x 2 − 3 )) d x = ∫ x 1 x 2 ( x − x 2 + 3 ) d x .
Evaluating the Integral Now we evaluate the definite integral: A re a = ∫ x 1 x 2 ( x − x 2 + 3 ) d x = [ 2 x 2 − 3 x 3 + 3 x ] x 1 x 2 . Plugging in the values of x 1 and x 2 , we get: A re a = ( 2 x 2 2 − 3 x 2 3 + 3 x 2 ) − ( 2 x 1 2 − 3 x 1 3 + 3 x 1 ) .
Calculating the Area Substituting x 1 = 2 1 − 13 and x 2 = 2 1 + 13 into the expression, we get: A re a = ( 2 ( 2 1 + 13 ) 2 − 3 ( 2 1 + 13 ) 3 + 3 ( 2 1 + 13 )) − ( 2 ( 2 1 − 13 ) 2 − 3 ( 2 1 − 13 ) 3 + 3 ( 2 1 − 13 )) . After simplifying, we find that A re a ≈ 7.812 .
Final Answer Therefore, the area between the parabola y = x 2 − 3 and the line y = x is approximately 7.812 square units.
Examples
Imagine you are designing a curved slide in a water park. The slide's path can be modeled by a parabolic function, and the straight ladder leading up to the slide can be modeled by a linear function. To determine how much material you need for the section between the slide and the ladder, you would calculate the area between these two curves, just like in this problem. This ensures you have enough material to build that specific section of the slide safely and efficiently.
