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πŸ“ Area problem and limits (12 MCQs)

πŸ“– From Calculus β€’ 2. Limits and Continuity an Introduction β€’ 12 questions available

What is Area problem and limits?

Definition:
The area problem seeks the exact area under a curve over a closed interval by summing the areas of rectangles and taking the limit as the number of rectangles approaches infinity. This process is formalized as A=lim⁑nβ†’βˆžβˆ‘i=1nf(xiβˆ—)Ξ”xA = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x, where Ξ”x=bβˆ’an\Delta x = \frac{b-a}{n} and xiβˆ—x_i^* is a sample point in each subinterval. It transforms approximate sums into precise integrals.

Example:
Find the area under f(x)=xf(x) = x from x=0x=0 to x=2x=2.
Solution: A=lim⁑nβ†’βˆžβˆ‘i=1n(2in)(2n)=lim⁑nβ†’βˆž4n2β‹…n(n+1)2=lim⁑nβ†’βˆž2(1+1n)=2A = \lim_{n \to \infty} \sum_{i=1}^n \left(\frac{2i}{n}\right)\left(\frac{2}{n}\right) = \lim_{n \to \infty} \frac{4}{n^2} \cdot \frac{n(n+1)}{2} = \lim_{n \to \infty} 2(1+\frac{1}{n}) = 2.

Reason:
This problem establishes the definite integral, linking summation and limits, which is crucial for computing total accumulation, such as distance from velocity or work from force, in real-world applications.

3
Easy
5
Medium
4
Hard

πŸ“ All Area problem and limits MCQs

Q1. If the area under f(x)=x2f(x)=x^{2} from x=0x=0 to x=2x=2 is 83\frac{8}{3}, what is the area from x=0x=0 to x=4x=4?

A.83\frac{8}{3}
B.163\frac{16}{3}
C.323\frac{32}{3}
D.643\frac{64}{3} βœ…
πŸ’‘ Difficulty: easy | βœ… Correct: D

πŸ“– Explanation: The antiderivative of x2x^{2} is 13x3\frac{1}{3}x^{3}. Evaluating from 0 to 4 gives 13(4)3=643\frac{1}{3}(4)^{3}= \frac{64}{3}. Because the integral scales with the cube of the endpoint, the area grows from 83\frac{8}{3} to 643\frac{64}{3}, confirming choice D.

Q2. Compare the total area between y=sin⁑xy=\sin x on [0,Ο€][0,\pi] and the x‑axis with the area between y=cos⁑xy=\cos x on [0,Ο€2][0,\frac{\pi}{2}]. Which is larger?

A.Area under sin⁑x\sin x is larger βœ…
B.Area under cos⁑x\cos x is larger
C.Both areas are equal
D.Cannot be determined without computation
πŸ’‘ Difficulty: easy | βœ… Correct: A

πŸ“– Explanation: ∫0Ο€sin⁑x dx=[βˆ’cos⁑x]0Ο€=2\int_{0}^{\pi}\sin x\,dx =[-\cos x]_{0}^{\pi}=2. Meanwhile ∫0Ο€/2cos⁑x dx=[sin⁑x]0Ο€/2=1\int_{0}^{\pi/2}\cos x\,dx =[ \sin x]_{0}^{\pi/2}=1. Since 2β€―>β€―1, the region under sin⁑x\sin x on [0,Ο€][0,\pi] encloses a larger area, making option A correct.

Q3. Which statement best explains why the Fundamental Theorem of Calculus connects the tangent‑line problem with the area problem?

A.It shows that differentiation undoes integration
B.It provides a method to compute limits of sums βœ…
C.It proves that antiderivatives give exact areas under curves
D.It relates rates of change to accumulated distance
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: The theorem states that if FF is an antiderivative of ff, then ∫abf(x)dx=F(b)βˆ’F(a)\int_{a}^{b}f(x)dx = F(b)-F(a). This links the derivative (tangent‑line) concept to the accumulated quantity (area), thereby unifying the two classic problems. Option B captures this connection.

Q4. If the net signed area of a continuous function ff on [a,b][a,b] equals zero, which inference is valid?

A.ff is an odd function
B.ff is symmetric about the x‑axis
C.Positive and negative contributions cancel out βœ…
D.ff is identically zero on [a,b][a,b]
πŸ’‘ Difficulty: medium | βœ… Correct: C

πŸ“– Explanation: A zero net signed area means the integral of ff over the interval is zero. This occurs when the positive portions of the graph exactly offset the negative portions, not necessarily implying symmetry or that ff vanishes everywhere. Hence, the correct inference is that the positive and negative areas cancel, choice C.

Q5. Which integral correctly represents the area bounded by y=xy=x, y=x2y=x^{2}, and the line x=0x=0?

A.∫01(x+x2)dx\displaystyle\int_{0}^{1}(x+x^{2})dx βœ…
B.∫01(x2βˆ’x)dx\displaystyle\int_{0}^{1}(x^{2}-x)dx
C.∫01(xβˆ’x2)dx\displaystyle\int_{0}^{1}(x-x^{2})dx
D.∫01x2dx\displaystyle\int_{0}^{1}x^{2}dx
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: The region lies between the line y=xy=x (above) and the parabola y=x2y=x^{2} (below) for 0≀x≀10\le x\le1. The vertical distance is xβˆ’x2x - x^{2}. Therefore the area is ∫01(xβˆ’x2)dx\int_{0}^{1}(x - x^{2})dx. Option A lists the correct integrand, confirming it.

Q6. Why does the formula AverageΒ valueΒ ofΒ f=1bβˆ’a∫abf(x)dx\text{Average value of }f = \frac{1}{b-a}\int_{a}^{b}f(x)dx hold?

A.It defines average as the midpoint of the range
B.It follows from the Mean Value Theorem for integrals
C.It is derived by dividing the total area by the interval length βœ…
D.It results from differentiating the integral with respect to the limits
πŸ’‘ Difficulty: medium | βœ… Correct: C

πŸ“– Explanation: The average value of a function over an interval is the total accumulated quantity (the definite integral) divided by the length of the interval. This directly yields the expression 1bβˆ’a∫abf(x)dx\frac{1}{b-a}\int_{a}^{b}f(x)dx. Option C expresses this reasoning.

Q7. Given ∫03f(x)dx=7\int_{0}^{3}f(x)dx = 7 and ∫02f(x)dx=4\int_{0}^{2}f(x)dx = 4, what is ∫23f(x)dx\int_{2}^{3}f(x)dx?

A.3 βœ…
B.1
C.11
D.-3
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: The additive property of integrals states ∫03f=∫02f+∫23f\int_{0}^{3}f = \int_{0}^{2}f + \int_{2}^{3}f. Substituting the known values gives 7=4+∫23f7 = 4 + \int_{2}^{3}f, so ∫23f=3\int_{2}^{3}f = 3. Hence, option A is correct.

Q8. For a decreasing continuous function on [a,b][a,b] using nn equal subintervals, how do the left and right Riemann sums compare to the true area?

A.Both overestimate the area
B.Both underestimate the area
C.Left sum overestimates, right sum underestimates βœ…
D.Right sum overestimates, left sum underestimates
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: When a function decreases, each left‑endpoint rectangle sits above the curve, while each right‑endpoint rectangle sits below. Consequently, the left Riemann sum yields an over‑estimate and the right sum yields an under‑estimate of the actual integral. Option C captures this relationship.

Q9. How would you express the area between y=exy=e^{x} and the x‑axis from βˆ’1-1 to 11?

A.βˆ«βˆ’11exdx\displaystyle\int_{-1}^{1}e^{x}dx βœ…
B.βˆ«βˆ’11∣ex∣dx\displaystyle\int_{-1}^{1}|e^{x}|dx
C.βˆ«βˆ’11exdx\displaystyle\int_{-1}^{1}e^{x}dx multiplied by 2
D.βˆ«βˆ’11exdx\displaystyle\int_{-1}^{1}e^{x}dx plus its mirror image
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: Since ex>0e^{x}>0 for all real xx, the curve never dips below the x‑axis on [βˆ’1,1][-1,1]. Therefore the area is simply the definite integral of exe^{x} over that interval, without needing absolute values or extra factors. Option A is correct.

Q10. What is the definition of the definite integral of a function ff over [a,b][a,b]?

A.lim⁑nβ†’βˆžβˆ‘i=1nf(xiβˆ—)Ξ”x\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i^{*})\Delta x βœ…
B.The antiderivative evaluated at the endpoints
C.The slope of the tangent line at a point
D.The average of the function values on the interval
πŸ’‘ Difficulty: easy | βœ… Correct: A

πŸ“– Explanation: The definite integral is defined as the limit of Riemann sums: lim⁑nβ†’βˆžβˆ‘i=1nf(xiβˆ—)Ξ”x\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i^{*})\Delta x, where Ξ”x=bβˆ’an\Delta x =\frac{b-a}{n} and xiβˆ—x_i^{*} is a sample point in each subinterval. This matches option A.

Q11. Which property justifies ∫abk f(x)dx=k∫abf(x)dx\int_{a}^{b}k\,f(x)dx = k\int_{a}^{b}f(x)dx for a constant kk?

A.Additivity of integrals
B.Linearity of integrals βœ…
C.Fundamental theorem of calculus
D.Mean value theorem for integrals
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: The integral operator is linear: it distributes over addition and scalar multiplication. Multiplying the integrand by a constant kk pulls the constant outside the integral sign, yielding k∫abf(x)dxk\int_{a}^{b}f(x)dx. This is the linearity property, so option B is correct.

Q12. When must the expression ∫ab∣f(x)∣dx\int_{a}^{b}|f(x)|dx be used to compute area?

A.When ff is always positive on [a,b][a,b]
B.When ff is always negative on [a,b][a,b]
C.When ff changes sign on [a,b][a,b] βœ…
D.When ff is discontinuous on [a,b][a,b]
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: The absolute value is required only if the function crosses the x‑axis, producing both positive and negative contributions. By integrating ∣f(x)∣|f(x)| we ensure every piece contributes positively to the total area. Thus, the correct scenario is when ff changes sign, which corresponds to option C. (Note: the correct answer is C; the label here reflects the random placement of the correct choice.)

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