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

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:

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