Fundamental Theorem practice problems

1. 1
   Find an antiderivative

   The derivative of x cubed is 3x squared.

   $$F(x)=x^3$$
2. 2
   Evaluate upper minus lower

   Apply the evaluation part of the theorem.

   $$F(2)-F(0)=8-0=8$$

Independent check: A right-endpoint Riemann sum approaches 8 as the number of subintervals increases.

1. 1
   Check the form

   The lower bound is constant and the upper bound is x.

   $$F(x)=\int_1^x f(t)\,dt$$
2. 2
   Apply Part 1

   The integrand is continuous for every real t.

   $$F'(x)=f(x)=\sqrt{1+x^2}$$

Independent check: A small increase h adds a strip whose average height approaches √(1 + x²).

1. 1
   Evaluate at the upper bound

   Part 1 gives cosine of the moving endpoint.

   $$\cos(x^3)$$
2. 2
   Apply the chain rule

   Differentiate x cubed.

   $$G'(x)=\cos(x^3)\cdot3x^2$$

Independent check: Since G(x) = sin(x³), direct differentiation gives the same result.

1. 1
   Reverse the bounds

   Changing orientation introduces a minus sign.

   $$H(x)=-\int_4^x\ln t\,dt$$
2. 2
   Differentiate

   Apply Part 1 to the reversed integral.

   $$H'(x)=-\ln x$$

Independent check: Differentiating an explicit antiderivative evaluation gives the same negative logarithm.

1. 1
   Differentiate the upper endpoint

   Evaluate t squared at x squared and multiply by 2x.

   $$(x^2)^2(2x)=2x^5$$
2. 2
   Differentiate the lower endpoint

   Subtract the integrand at x times the derivative of x.

   $$x^2(1)=x^2$$
3. 3
   Combine

   Keep the orientation signs.

   $$K'(x)=2x^5-x^2$$

Independent check: Evaluating the integral first gives x⁶/3 - x³/3, whose derivative is 2x⁵ - x².

1. 1
   Choose an antiderivative

   On the positive interval, the reciprocal integrates to natural logarithm.

   $$F(x)=\ln x$$
2. 2
   Evaluate the endpoints

   Use upper minus lower.

   $$\ln e-\ln1=1-0=1$$

Independent check: The substitution x = e^u changes the integral to the length of the u-interval from 0 to 1.

1. 1
   Find an antiderivative

   Integrate the linear function.

   $$t^2-3t$$
2. 2
   Evaluate at 4 and 0

   Apply the evaluation theorem.

   $$(16-12)-0=4$$

Independent check: Geometrically, the negative triangle has area 9/4 and the positive triangle has area 25/4, leaving 4.

1. 1
   Use the average-value formula

   The interval length is 2.

   $$f_{\mathrm{avg}}=\frac12\int_0^2x^2\,dx$$
2. 2
   Evaluate

   The integral is 8/3.

   $$f_{\mathrm{avg}}=\frac12\cdot\frac83=\frac43$$

Independent check: One-panel Simpson's rule is exact for x²: (2/6)[0 + 4(1) + 4] = 8/3; dividing by the interval length 2 gives 4/3.

1. 1
   Locate the sign change

   The graph crosses the axis at zero.

   $$x=0$$
2. 2
   Add positive areas

   Each triangle has base 1 and height 1.

   $$A=\frac12+\frac12=1$$

Independent check: Directly integrating |x| from -1 to 1 gives 1.

1. 1
   Check the integrand

   The left and right values at zero differ.

   $$\lim_{t\to0^-}f(t)=1,\qquad\lim_{t\to0^+}f(t)=2$$
2. 2
   Translate to accumulation slopes

   A gains area at rate 1 from the left and rate 2 from the right.

   $$A'_-(0)=1,\qquad A'_+(0)=2$$
3. 3
   Conclude

   Unequal one-sided derivatives mean A prime does not exist at zero.

   $$A'(0)\text{ does not exist}$$

Independent check: Writing A piecewise produces two line segments with different slopes that meet at x = 0.