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Derivative and Definite Integral Application

Created by @nathanedwards
 at October 31st 2023, 10:22:58 pm.
AP_Calculus_AB-Questions
#calculus
#fundamental-theorem-of-calculus
#derivative

AP Calculus AB Exam Question:

Let f(x) be a continuous function on the interval [a, b] given by the equation:

f(x)=x32−x2+2x−3f(x) = \frac{x^3}{2} - x^2 + 2x - 3(a)
F(x)=∫axf(t)dtF(x) = \int_{a}^{x} f(t) dt(b)

Answer:

(a) To find the derivative of F(x), apply the Fundamental Theorem of Calculus:

According to the theorem, if F(x) is the antiderivative of f(x), then:

F′(x)=f(x)F'(x) = f(x)

Therefore, to find F'(x), we need to find the antiderivative of f(x) by integrating the function. Let's proceed with the integration:

F(x)=∫axf(t)dtF(x) = \int_{a}^{x} f(t) dt
F(x)=∫ax(t32−t2+2t−3)dtF(x) = \int_{a}^{x} \left( \frac{t^3}{2} - t^2 + 2t - 3 \right) dt

Using the power rule and the properties of integration, we can integrate term by term:

F(x)=12∫axt3dt−∫axt2dt+2∫axtdt−3∫axdtF(x) = \frac{1}{2} \int_{a}^{x} t^3 dt - \int_{a}^{x} t^2 dt + 2 \int_{a}^{x} t dt - 3 \int_{a}^{x} dt
F(x)=12[t44]ax−[t33]ax+2[t22]ax−3[t]axF(x) = \frac{1}{2} \left[\frac{t^4}{4}\right]_{a}^{x} - \left[\frac{t^3}{3}\right]_{a}^{x} + 2 \left[\frac{t^2}{2}\right]_{a}^{x} - 3 \left[t\right]_{a}^{x}

Simplifying further:

F(x)=18x4−13x3+x2−32x−18a4+13a3−a2+32aF(x) = \frac{1}{8}x^4 - \frac{1}{3}x^3 + x^2 - \frac{3}{2}x - \frac{1}{8}a^4 + \frac{1}{3}a^3 - a^2 + \frac{3}{2}a

Therefore, the derivative of F(x), based on the Fundamental Theorem of Calculus, is:

F′(x)=ddx[18x4−13x3+x2−32x−18a4+13a3−a2+32a]F'(x) = \frac{d}{dx} \left[\frac{1}{8}x^4 - \frac{1}{3}x^3 + x^2 - \frac{3}{2}x - \frac{1}{8}a^4 + \frac{1}{3}a^3 - a^2 + \frac{3}{2}a\right]
F′(x)=ddx[18x4−13x3+x2−32x](constant terms do not affect the derivative)F'(x) = \frac{d}{dx} \left[\frac{1}{8}x^4 - \frac{1}{3}x^3 + x^2 - \frac{3}{2}x\right] \quad \text{(constant terms do not affect the derivative)}
F′(x)=18⋅4x3−13⋅3x2+2x−32F'(x) = \frac{1}{8}\cdot4x^3 - \frac{1}{3}\cdot3x^2 + 2x - \frac{3}{2}

Simplifying further:

F′(x)=12x3−x2+2x−32F'(x) = \frac{1}{2}x^3 - x^2 + 2x - \frac{3}{2}(b)
F(b)−F(a)=(18b4−13b3+b2−32b)−(18a4−13a3+a2−32a)F(b) - F(a) = \left(\frac{1}{8}b^4 - \frac{1}{3}b^3 + b^2 - \frac{3}{2}b\right) - \left(\frac{1}{8}a^4 - \frac{1}{3}a^3 + a^2 - \frac{3}{2}a\right)

Simplifying further:

F(b)−F(a)=18b4−18a4−13b3+13a3+b2−a2−32b+32aF(b) - F(a) = \frac{1}{8}b^4 - \frac{1}{8}a^4 - \frac{1}{3}b^3 + \frac{1}{3}a^3 + b^2 - a^2 - \frac{3}{2}b + \frac{3}{2}a

Therefore, F(b) - F(a) can be written as:

F(b)−F(a)=18(b4−a4)−13(b3−a3)+(b2−a2)−32(b−a)F(b) - F(a) = \frac{1}{8}(b^4 - a^4) - \frac{1}{3}(b^3 - a^3) + (b^2 - a^2) - \frac{3}{2}(b - a)

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