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  1. AP Calculus
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Applications of Integration

Question 1
college-boardCalculus AB/BCAPExam Style
1 mark

Given that two functions intersect at points (a,b)(a,b)(a,b) and (c,d)(c,d)(c,d) with a<ca < ca<c, which integral represents finding area between these curves from leftmost to rightmost intersection point?

Question 2
college-boardCalculus AB/BCAPExam Style
1 mark

Find the area between the curves y=e2xy = e^{2x}y=e2x and y=2exy = 2e^xy=2ex for 0leqxleqln(2)0 \\leq x \\leq \\ln(2)0leqxleqln(2).

Question 3
college-boardCalculus AB/BCAPExam Style
1 mark

Find the area between the curves y=x2y = x^2y=x2 and y=2xy = 2xy=2x for 0≤x≤20 \leq x \leq 20≤x≤2.

Question 4
college-boardCalculus AB/BCAPExam Style
1 mark

Find the area between the curves y=exy = e^xy=ex and y=xy = xy=x for 0≤x≤10 \leq x \leq 10≤x≤1.

Question 5
college-boardCalculus AB/BCAPExam Style
1 mark

What value of k would make continuous on [-1, 1] if f(x)={k+xfor x<0sin⁡(πkx)for x≥0f(x) = \begin{cases} k + x & \text{for } x < 0 \\ \sin\left(\frac{\pi}{k} x\right) & \text{for } x \geq 0 \end{cases}f(x)={k+xsin(kπ​x)​for x<0for x≥0​?

Question 6
college-boardCalculus AB/BCAPExam Style
1 mark

Determine the area enclosed by the curves y=sin⁡(x)y = \sin(x)y=sin(x) and y=cos⁡(x)y = \cos(x)y=cos(x) for 0≤x≤π20 \leq x \leq \frac{\pi}{2}0≤x≤2π​.

Question 7
college-boardCalculus AB/BCAPExam Style
1 mark

Determine the area enclosed by the curves y=x3y = x^3y=x3 and y=3xy = 3xy=3x for −1leqxleq1-1 \\leq x \\leq 1−1leqxleq1.

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Question 8
college-boardCalculus AB/BCAPExam Style
1 mark

If you set up an integral to find the area between x\sqrt{x}x​ and x2\frac{x}{2}2x​ from x=1x=1x=1 to x=4x=4x=4, which function should be on top in your integral setup?

Question 9
college-boardCalculus AB/BCAPExam Style
1 mark

What is the area of the region enclosed by y=x2y = x^2y=x2 and y=2xy = 2xy=2x, between x=0x=0x=0 and their intersection point?

Question 10
college-boardCalculus AB/BCAPExam Style
1 mark

Determine the area enclosed by the curves y=x2y = x^2y=x2 and y=xy = \sqrt{x}y=x​ for 0≤x≤10 \leq x \leq 10≤x≤1.