Find the area enclosed between the straight line y=12x+14 and the curve y=3x^2+6x+5
The graphs intersect at (-1,2) and (3,50)
So, consider the area as a collection of thin strips, whose height is the distance between the curves. That makes the area
∫[-1,3] (12x+14)-(3x^2+6x+5) dx = ∫[-1,3] -3x^2 + 6x + 9 dx
= 3∫[-1,3] -x^2 + 2x + 3 dx = 32
To find the area enclosed between the straight line and the curve, we need to determine the points of intersection first.
Step 1: Set the two equations equal to each other:
12x + 14 = 3x^2 + 6x + 5
Step 2: Rearrange the equation to form a quadratic equation:
3x^2 - 6x + 5 = -12x - 14
Step 3: Combine like terms:
3x^2 - 6x - 12x + 5 + 14 = 0
3x^2 - 18x + 19 = 0
Step 4: Solve the quadratic equation using factoring, completing the square, or using the quadratic formula:
Since this equation doesn't factor nicely, we can use the quadratic formula to find the values of x.
The quadratic formula states that for an equation in the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 3, b = -18, and c = 19. Plugging these values into the quadratic formula, we get:
x = (-(-18) ± √((-18)^2 - 4 * 3 * 19)) / (2 * 3)
x = (18 ± √(324 - 228)) / 6
x = (18 ± √96) / 6
x = (18 ± 4√6) / 6
x = 3 ± √6
So, the two points of intersection are x = 3 + √6 and x = 3 - √6.
Step 5: To find the corresponding y-values, substitute the x-values into either of the original equations:
For y = 12x + 14:
y = 12(3 + √6) + 14
y = 36 + 12√6 + 14
y = 50 + 12√6
y = 12(3 - √6) + 14
y = 36 - 12√6 + 14
y = 50 - 12√6
So, the two points of intersection are (3 + √6, 50 + 12√6) and (3 - √6, 50 - 12√6).
Step 6: To find the area enclosed, we integrate the curve function minus the line function over the interval between the two x-values of intersection.
The area between two curves can be calculated as follows:
Area = ∫[a, b] (f(x) - g(x)) dx
where f(x) and g(x) are the two curves, and [a, b] is the interval between the two x-values of intersection.
In this case, we have:
Area = ∫[(3 - √6), (3 + √6)] [(3x^2 + 6x + 5) - (12x + 14)] dx
Now, we can simplify the equation and integrate it:
Area = ∫[(3 - √6), (3 + √6)] (3x^2 - 6x + 5 - 12x - 14) dx
Area = ∫[(3 - √6), (3 + √6)] (3x^2 - 18x - 9) dx
To integrate this equation, we use the power rule for integration.
The integral of x^n with respect to x is given by:
∫x^n dx = (1/(n+1)) * x^(n+1) + C
Applying the power rule:
Area = [(1/3) * x^3 - 9 * x^2 - 9x] evaluated from (3 - √6) to (3 + √6)
Evaluating the integral:
Area = [(1/3) * (3 + √6)^3 - 9 * (3 + √6)^2 - 9 * (3 + √6)] - [(1/3) * (3 - √6)^3 - 9 * (3 - √6)^2 - 9 * (3 - √6)]
Area = [(1/3) * (27 + 27√6 + 18 + 6√6 + 6 + √6) - 9 * (9 + 18√6 + 6) - 9 * (3 + √6)] - [(1/3) * (27 - 27√6 + 18 - 6√6 + 6 - √6) - 9 * (9 - 18√6 + 6) - 9 * (3 - √6)]
Simplifying the expression:
Area = [(1/3) * (51 + 34√6) - (9 * (33 + 18√6) + 9 * (3 + √6))] - [(1/3) * (51 - 34√6) - (9 * (33 - 18√6) + 9 * (3 - √6))]
Area = 2√6 + 50
Therefore, the area enclosed between the straight line y = 12x + 14 and the curve y = 3x^2 + 6x + 5 is 2√6 + 50 square units.