Question

bc unit 6 (all topics) quiz 2025
(a) ⅓(b³ - a³)
(b) b³ - a³
(c) ¼(b⁴ - a⁴)
(d) b⁴ - a⁴
(e) 2(b⁴ - a⁴)

1. which of the following is an antiderivative of f(x)=√(1 + x³)?

(a) ⅔(1 + x⁸)⅓
(b) ⅓(1 + x³)⅔/3x²
(c) ∫₀^(1 + x⁸)√t dt
(d) ∫₀^(x³)√(1 + t) dt
(e) ∫₀^x√(1 + t³) dt

1. let f be the function defined by f(x)= - 2x+∫₂^x√(1 + t²) dt. which of the following statements is true on the interval 0 < x < 1?

(a) f is decreasing, and the graph of f is concave up.
(b) f is decreasing, and the graph of f is concave down.
(c) f is increasing, and the graph of f is concave up.
(d) f is increasing, and the graph of f is concave down.

1. the graph of a function f is shown above. what is the value of ∫₀⁷f(x) dx?

Explanation:

Step 1: Analyze question 4

By the fundamental theorem of calculus, if \(F(x)=\int_{a}^{x}f(t)dt\), then \(F^\prime(x) = f(x)\). An antiderivative of \(f(x)=\sqrt{1 + x^{3}}\) is of the form \(\int_{0}^{x}\sqrt{1 + t^{3}}dt+C\). When \(C = 0\), \(\int_{0}^{x}\sqrt{1 + t^{3}}dt\) is an antiderivative. So the answer to question 4 is E.

Step 2: Analyze question 5

First, find the derivative of \(f(x)=-2x+\int_{2}^{x}\sqrt{1 + t^{2}}dt\). Using the sum - rule and the fundamental theorem of calculus, \(f^\prime(x)=-2+\sqrt{1 + x^{2}}\). For \(0\lt x\lt1\), \(\sqrt{1 + x^{2}}\in(1,\sqrt{2})\), so \(f^\prime(x)=-2+\sqrt{1 + x^{2}}\lt0\), which means \(f\) is decreasing. Then find the second - derivative \(f^{\prime\prime}(x)=\frac{x}{\sqrt{1 + x^{2}}}\). For \(0\lt x\lt1\), \(f^{\prime\prime}(x)>0\), so the graph of \(f\) is concave up. The answer to question 5 is A.

Step 3: Analyze question 6

We use the geometric interpretation of the definite integral \(\int_{0}^{7}f(x)dx\) as the net area between the curve \(y = f(x)\) and the \(x\) - axis.
The area of the trapezoid from \(x = 0\) to \(x = 2\) is \(A_1=\frac{(1 + 2)}{2}\times2=3\).
The area of the rectangle from \(x = 2\) to \(x = 4\) is \(A_2=2\times2 = 4\).
The area of the triangle from \(x = 4\) to \(x = 5\) is \(A_3=\frac{1}{2}\times2\times1 = 1\).
The area of the triangle from \(x = 5\) to \(x = 7\) is \(A_4=-\frac{1}{2}\times2\times2=- 2\).
Then \(\int_{0}^{7}f(x)dx=A_1 + A_2+A_3+A_4=3 + 4+1-2=6\).

Answer:

1. E. \(\int_{0}^{x}\sqrt{1 + t^{3}}dt\)
2. A. \(f\) is decreasing, and the graph of \(f\) is concave up.
3. 6