Math  /  Calculus

Question5. Prove that the functions F1(x)=sin2(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
To find the constant difference, consider F1(x)F2(x) F_1(x) - F_2(x) .
F1(x)F2(x)=(sin2(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)=sin2(x)+7+12cos(2x)+11 F_1(x) - F_2(x) = \sin^2(x) + 7 + \frac{1}{2} \cos(2x) + 11
Using the identity sin2(x)=1cos(2x)2 \sin^2(x) = \frac{1 - \cos(2x)}{2} : F1(x)F2(x)=1cos(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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