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Math  /  Calculus

Question5. Prove that the functions F1(x)=sin⁡2(x)+7F_{1}(x)=\sin ^{2}(x)+7 and F2(x)=−12cos⁡(2x)−11F_{2}(x)=-\frac{1}{2} \cos (2 x)-11 are antiderivatives of the same function ff. Give a formula for ff at xx. Compute the numerical value of the constant by which the two antiderivatives F1F_{1} and F2F_{2} differ.

Studdy Solution

STEP 1

1. F1(x)=sin⁡2(x)+7 F_1(x) = \sin^2(x) + 7 and F2(x)=−12cos⁡(2x)−11 F_2(x) = -\frac{1}{2} \cos(2x) - 11 are antiderivatives of the same function f(x) f(x) .
2. We need to find the function f(x) f(x) and the constant difference between F1 F_1 and F2 F_2 .

STEP 2

1. Differentiate F1(x) F_1(x) to find f(x) f(x) .
2. Differentiate F2(x) F_2(x) to find f(x) f(x) .
3. Compare the derivatives to verify they are the same.
4. Calculate the constant difference between F1(x) F_1(x) and F2(x) F_2(x) .

STEP 3

Differentiate F1(x)=sin⁡2(x)+7 F_1(x) = \sin^2(x) + 7 .
Using the chain rule, the derivative of sin⁡2(x) \sin^2(x) is 2sin⁡(x)cos⁡(x)=sin⁡(2x) 2\sin(x)\cos(x) = \sin(2x) .
Thus, F1′(x)=sin⁡(2x) F_1'(x) = \sin(2x) .

STEP 4

Differentiate F2(x)=−12cos⁡(2x)−11 F_2(x) = -\frac{1}{2} \cos(2x) - 11 .
The derivative of −12cos⁡(2x) -\frac{1}{2} \cos(2x) is sin⁡(2x) \sin(2x) (using the chain rule and the derivative of cosine).
Thus, F2′(x)=sin⁡(2x) F_2'(x) = \sin(2x) .

STEP 5

Since F1′(x)=sin⁡(2x) F_1'(x) = \sin(2x) and F2′(x)=sin⁡(2x) F_2'(x) = \sin(2x) , both derivatives are the same, confirming that both F1(x) F_1(x) and F2(x) F_2(x) are antiderivatives of the same function f(x)=sin⁡(2x) f(x) = \sin(2x) .

STEP 6

To find the constant difference, consider F1(x)−F2(x) F_1(x) - F_2(x) .
F1(x)−F2(x)=(sin⁡2(x)+7)−(−12cos⁡(2x)−11) F_1(x) - F_2(x) = (\sin^2(x) + 7) - \left(-\frac{1}{2} \cos(2x) - 11\right)
Simplify: F1(x)−F2(x)=sin⁡2(x)+7+12cos⁡(2x)+11 F_1(x) - F_2(x) = \sin^2(x) + 7 + \frac{1}{2} \cos(2x) + 11
Using the identity sin⁡2(x)=1−cos⁡(2x)2 \sin^2(x) = \frac{1 - \cos(2x)}{2} : F1(x)−F2(x)=1−cos⁡(2x)2+7+12cos⁡(2x)+11 F_1(x) - F_2(x) = \frac{1 - \cos(2x)}{2} + 7 + \frac{1}{2} \cos(2x) + 11
Simplify further: F1(x)−F2(x)=12+7+11=18.5 F_1(x) - F_2(x) = \frac{1}{2} + 7 + 11 = 18.5
Thus, the constant by which the two antiderivatives differ is 18.5 18.5 .
The function f(x) f(x) is sin⁡(2x) \sin(2x) and the constant difference is:
18.5 \boxed{18.5}

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