Question
If \(f(x)=\int _{0}^{x^{3}}\cos (t^{2})dt\) , then \(f'(\sqrt{\pi })=\)
A \(3π sin(\pi ^{3})\)
B \(cos(\pi ^{3})\)
C 3π cosπ
D 3π \(cos(\pi ^{3})\)
▶️ Answer/Explanation
Solution
1. Apply the Fundamental Theorem of Calculus Part 1 and Chain Rule:
For \( f(x) = \int_{0}^{x^{3}} \cos(t^{2}) dt \), the derivative is: \( f'(x) = \cos((x^3)^2) \cdot \frac{d}{dx}(x^3) \)
2. Differentiate the upper limit:
\( f'(x) = \cos(x^6) \cdot 3x^2 \)
3. Evaluate at \( x = \sqrt{\pi} \):
\( f'(\sqrt{\pi}) = \cos((\sqrt{\pi})^6) \cdot 3(\sqrt{\pi})^2 \)
\( = \cos(\pi^3) \cdot 3\pi \)
\( = 3\pi \cos(\pi^3) \)
4. Verify against options:
The correct answer matches option D.
✅ Answer: D) \( 3π \cos(\pi^{3}) \)
Note: This problem combines the Fundamental Theorem of Calculus with the Chain Rule. The key steps are:
- Differentiating the integral with variable upper limit
- Applying the chain rule for the composite function \( x^3 \)
- Carefully evaluating at \( x = \sqrt{\pi} \)
Question
\(\int_{-1}^{2}\frac{|x|}{x}dx\) is
(A) –3 (B) 1 (C) 2 (D) 3 (E) nonexistent
▶️ Answer/Explanation
Solution
1. Analyze the integrand \(\frac{|x|}{x}\):
• For \(x > 0\): \(\frac{|x|}{x} = 1\)
• For \(x < 0\): \(\frac{|x|}{x} = -1\)
• At \(x = 0\): Undefined, but integrable with finite discontinuity
• For \(x < 0\): \(\frac{|x|}{x} = -1\)
• At \(x = 0\): Undefined, but integrable with finite discontinuity
2. Split the integral at \(x = 0\):
\(\int_{-1}^{2} \frac{|x|}{x} dx = \int_{-1}^{0} -1 dx + \int_{0}^{2} 1 dx\)
3. Evaluate each integral separately:
• \(\int_{-1}^{0} -1 dx = -1 \times (0 – (-1)) = -1\)
• \(\int_{0}^{2} 1 dx = 1 \times (2 – 0) = 2\)
• \(\int_{0}^{2} 1 dx = 1 \times (2 – 0) = 2\)
4. Sum the results:
\(-1 + 2 = 1\)
✅ Answer: B) 1
Question
Let \(f(t)=t^{4}+2t^{2}-3\) and consider the function \(F(x)=\int_{-3}^{x}f(t)dt\) What are the x-value(s) for which F(x) has a local minimum?
(A) 1
(B) -1
(C) \(\pm \sqrt{3}\)
(D) ±1
▶️ Answer/Explanation
Solution
1. Find critical points using the Fundamental Theorem of Calculus:
\( F'(x) = f(x) = x^{4} + 2x^{2} – 3 \)
Set \( F'(x) = 0 \): \( x^{4} + 2x^{2} – 3 = 0 \)
2. Solve the quartic equation:
Let \( u = x^{2} \): \( u^{2} + 2u – 3 = 0 \)
Solutions: \( u = \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm 4}{2} \)
\( u = 1 \) or \( u = -3 \) (discard negative solution)
Thus \( x^{2} = 1 \) ⇒ \( x = \pm 1 \)
3. Determine local minima using second derivative test:
\( F”(x) = f'(x) = 4x^{3} + 4x \)
At \( x = 1 \): \( F”(1) = 4 + 4 = 8 > 0 \) ⇒ local minimum
At \( x = -1 \): \( F”(-1) = -4 -4 = -8 < 0 \) ⇒ local maximum
4. Verify behavior:
The only local minimum occurs at \( x = 1 \)
✅ Answer: A) 1
Note: While both x=1 and x=-1 are critical points, only x=1 satisfies the second derivative test for a local minimum. The function changes from decreasing to increasing at x=1.
Question
If \(g(x)=\int_{0}^{x}\left ( \log _{2} (t-1)+\log _{2}(t+1)\right )dt\) find all x-values at which the line tangent to the graph of y = g(x) is horizontal.
(A) \(\sqrt{2}\)
(B) \(\sqrt{3}\)
(C) 2
(D) 3
▶️ Answer/Explanation
Solution
1. Find where the tangent is horizontal (g'(x) = 0):
By the Fundamental Theorem of Calculus: \( g'(x) = \log_{2}(x-1) + \log_{2}(x+1) \)
2. Combine logarithms using product rule:
\( \log_{2}(x-1) + \log_{2}(x+1) = \log_{2}[(x-1)(x+1)] = 0 \)
3. Convert logarithmic equation to exponential form:
\( (x-1)(x+1) = 2^0 = 1 \)
\( x^2 – 1 = 1 \)
\( x^2 = 2 \)
\( x = \pm \sqrt{2} \)
4. Check domain restrictions:
• \( \log_{2}(x-1) \) requires \( x > 1 \)
• \( \log_{2}(x+1) \) requires \( x > -1 \)
Thus \( x > 1 \), eliminating \( x = -\sqrt{2} \)
5. Verify \( x = \sqrt{2} \) is within integration limits:
The lower limit 0 is valid since the integrand is only defined for t > 1
✅ Answer: A) \( \sqrt{2} \)
Note: The only valid solution is \( x = \sqrt{2} \) because:
- It satisfies g'(x) = 0
- It’s within the domain of both logarithmic functions (x > 1)
- The integral is properly defined from 0 to \( \sqrt{2} \) since the integrand is defined for t > 1