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  2. AP Calculus
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Explain the core concept of the Fundamental Theorem of Calculus.

It establishes the relationship between differentiation and integration, showing that they are inverse operations.

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Explain the core concept of the Fundamental Theorem of Calculus.

It establishes the relationship between differentiation and integration, showing that they are inverse operations.

Explain how the Fundamental Theorem of Calculus is used to find derivatives of integrals.

It states that the derivative of an integral with a variable upper limit is simply the integrand evaluated at that upper limit.

What happens when the upper bound of the integral is a function of x?

You must apply the chain rule by multiplying the integrand evaluated at that function by the derivative of the function.

What is the significance of the Fundamental Theorem of Calculus?

It provides a method to evaluate definite integrals and connects the concepts of area and rate of change.

How do you handle a constant of integration when using the Fundamental Theorem of Calculus?

For definite integrals, the constant of integration cancels out when evaluating at the upper and lower limits.

Explain the relationship between the area under a curve and the antiderivative.

The definite integral, calculated using the antiderivative, gives the net area under the curve between the limits of integration.

Why is continuity important for the Fundamental Theorem of Calculus?

The Fundamental Theorem of Calculus requires the function to be continuous on the interval to ensure the existence of the integral and derivative.

What does the Fundamental Theorem of Calculus tell us about displacement and velocity?

The integral of velocity gives displacement, showing how integration can be used to find position from velocity.

Explain how the Fundamental Theorem of Calculus simplifies finding areas.

It provides a direct method to calculate the area under a curve by finding the antiderivative and evaluating at the bounds.

What are some common mistakes when applying the Fundamental Theorem of Calculus?

Forgetting the chain rule when the upper limit is a function, or not switching the bounds when needed.

How to find g′(x)g'(x)g′(x) if g(x)=∫axf(t)dtg(x) = \int_{a}^{x} f(t) dtg(x)=∫ax​f(t)dt?

Apply the Fundamental Theorem of Calculus: g′(x)=f(x)g'(x) = f(x)g′(x)=f(x).

How to find g′(x)g'(x)g′(x) if g(x)=∫au(x)f(t)dtg(x) = \int_{a}^{u(x)} f(t) dtg(x)=∫au(x)​f(t)dt?

Apply the Fundamental Theorem of Calculus and the Chain Rule: g′(x)=f(u(x))cdotu′(x)g'(x) = f(u(x)) cdot u'(x)g′(x)=f(u(x))cdotu′(x).

How to find g′(c)g'(c)g′(c) given g(x)=∫axf(t)dtg(x) = \int_{a}^{x} f(t) dtg(x)=∫ax​f(t)dt?

First find g′(x)g'(x)g′(x) using the Fundamental Theorem of Calculus, then evaluate g′(c)g'(c)g′(c).

How to find F′(x)F'(x)F′(x) if F(x)=∫u(x)bf(t)dtF(x) = \int_{u(x)}^{b} f(t) dtF(x)=∫u(x)b​f(t)dt?

First, switch the limits: F(x)=−∫bu(x)f(t)dtF(x) = -\int_{b}^{u(x)} f(t) dtF(x)=−∫bu(x)​f(t)dt. Then, F′(x)=−f(u(x))cdotu′(x)F'(x) = -f(u(x)) cdot u'(x)F′(x)=−f(u(x))cdotu′(x).

How to find g′(x)g'(x)g′(x) if g(x)=∫u(x)v(x)f(t)dtg(x) = \int_{u(x)}^{v(x)} f(t) dtg(x)=∫u(x)v(x)​f(t)dt?

Split the integral: g(x)=∫0v(x)f(t)dt−∫0u(x)f(t)dtg(x) = \int_{0}^{v(x)} f(t) dt - \int_{0}^{u(x)} f(t) dtg(x)=∫0v(x)​f(t)dt−∫0u(x)​f(t)dt. Then, g′(x)=f(v(x))v′(x)−f(u(x))u′(x)g'(x) = f(v(x))v'(x) - f(u(x))u'(x)g′(x)=f(v(x))v′(x)−f(u(x))u′(x).

How do you evaluate ddx∫2x3cos⁡(t)dt\frac{d}{dx} \int_{2}^{x^3} \cos(t) dtdxd​∫2x3​cos(t)dt?

Apply the Fundamental Theorem of Calculus and the chain rule: cos⁡(x3)cdot3x2\cos(x^3) cdot 3x^2cos(x3)cdot3x2.

How do you find F′(x)F'(x)F′(x) if F(x)=∫5x(t2+1)dtF(x) = \int_{5}^{x} (t^2 + 1) dtF(x)=∫5x​(t2+1)dt?

Apply the Fundamental Theorem of Calculus: F′(x)=x2+1F'(x) = x^2 + 1F′(x)=x2+1.

How do you solve for g′(2)g'(2)g′(2) if g(x)=∫1x2tdtg(x) = \int_{1}^{x^2} \sqrt{t} dtg(x)=∫1x2​t​dt?

First find g′(x)=x2cdot2x=∣x∣cdot2xg'(x) = \sqrt{x^2} cdot 2x = |x| cdot 2xg′(x)=x2​cdot2x=∣x∣cdot2x. Then, g′(2)=22cdot2(2)=8g'(2) = \sqrt{2^2} cdot 2(2) = 8g′(2)=22​cdot2(2)=8.

How do you find the derivative of ∫x0et2dt\int_{x}^{0} e^{t^2} dt∫x0​et2dt?

Switch the bounds and apply the Fundamental Theorem of Calculus: −∫0xet2dt-\int_{0}^{x} e^{t^2} dt−∫0x​et2dt, so the derivative is −ex2-e^{x^2}−ex2.

How do you find h′(x)h'(x)h′(x) if h(x)=∫sin⁡(x)3t3dth(x) = \int_{\sin(x)}^{3} t^3 dth(x)=∫sin(x)3​t3dt?

Switch the bounds: h(x)=−∫3sin⁡(x)t3dth(x) = -\int_{3}^{\sin(x)} t^3 dth(x)=−∫3sin(x)​t3dt. Apply the Fundamental Theorem of Calculus and the chain rule: h′(x)=−(sin⁡(x))3cos⁡(x)h'(x) = -(\sin(x))^3 \cos(x)h′(x)=−(sin(x))3cos(x).

What is the formula for the Fundamental Theorem of Calculus (Part 1)?

ddxleft[intaxf(t)dtight]=f(x)\frac{d}{dx}left[int_{a}^{x}f(t)dt ight]=f(x)dxd​left[intax​f(t)dtight]=f(x)

What is the formula for switching the bounds of integration?

∫abf(x),dx=−∫baf(x),dx\int_{a}^{b}f(x) , dx = -\int_{b}^{a}f(x) , dx∫ab​f(x),dx=−∫ba​f(x),dx

What is the formula for the Fundamental Theorem of Calculus (Part 1) with a function in the upper limit?

ddxleft[intag(x)f(t)dtight]=f(g(x))cdotg′(x)\frac{d}{dx}left[int_{a}^{g(x)}f(t)dt ight]=f(g(x)) cdot g'(x)dxd​left[intag(x)​f(t)dtight]=f(g(x))cdotg′(x)

What is the general form of an accumulation function?

F(x)=∫axf(t),dtF(x) = \int_{a}^{x} f(t) , dtF(x)=∫ax​f(t),dt