QuestionJuly 28, 2025

Find the area between the two curves y=2x and y=x^2-15 from x=-3 to x=3 The area between the curves is square fType an integer or a fraction.)

Find the area between the two curves y=2x and y=x^2-15 from x=-3 to x=3 The area between the curves is square fType an integer or a fraction.)
Find the area between the two curves y=2x and y=x^2-15 from x=-3 to x=3 The area between the curves is square fType an integer or a fraction.)

Solution
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Answer

72 Explanation 1. Identify Intersection Points Set 2x = x^2 - 15. Solve x^2 - 2x - 15 = 0 using the quadratic formula: x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=1, b=-2, c=-15. This gives x = 5 and x = -3. 2. Determine Integration Limits The limits are from x = -3 to x = 3 as given in the problem. 3. Setup Integral for Area The area is \int_{-3}^{3} [(2x) - (x^2 - 15)] \, dx = \int_{-3}^{3} (2x - x^2 + 15) \, dx. 4. Integrate Compute \int (2x - x^2 + 15) \, dx = [x^2 - \frac{x^3}{3} + 15x]. 5. Evaluate Definite Integral Evaluate from -3 to 3: At x = 3: (3)^2 - \frac{(3)^3}{3} + 15(3) = 9 - 9 + 45 = 45. At x = -3: (-3)^2 - \frac{(-3)^3}{3} + 15(-3) = 9 + 9 - 45 = -27. Subtract: 45 - (-27) = 72.

Explanation

1. Identify Intersection Points<br /> Set $2x = x^2 - 15$. Solve $x^2 - 2x - 15 = 0$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=-2$, $c=-15$. This gives $x = 5$ and $x = -3$.<br />2. Determine Integration Limits<br /> The limits are from $x = -3$ to $x = 3$ as given in the problem.<br />3. Setup Integral for Area<br /> The area is $\int_{-3}^{3} [(2x) - (x^2 - 15)] \, dx = \int_{-3}^{3} (2x - x^2 + 15) \, dx$.<br />4. Integrate<br /> Compute $\int (2x - x^2 + 15) \, dx = [x^2 - \frac{x^3}{3} + 15x]$.<br />5. Evaluate Definite Integral<br /> Evaluate from $-3$ to $3$: <br /> At $x = 3$: $(3)^2 - \frac{(3)^3}{3} + 15(3) = 9 - 9 + 45 = 45$.<br /> At $x = -3$: $(-3)^2 - \frac{(-3)^3}{3} + 15(-3) = 9 + 9 - 45 = -27$.<br /> Subtract: $45 - (-27) = 72$.
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