Q: The additivity property of definite integrals states \int_a^b f\,dx =

\int_a^c f\,dx + \int_c^b f\,dx. An interval can be split at any interior point c: the integral over [a,b] is the sum of the parts.

Question 1

The Fundamental Theorem of Calculus states \int_a^b f(x)\,dx =

Accepted Answer

F(b) - F(a), where F is an antiderivative of f. Evaluate any antiderivative F at the upper limit minus the lower limit: F(b) − F(a).

Question 2

Geometrically, the definite integral \int_a^b f(x)\,dx represents:

Accepted Answer

The (signed) area between the curve y = f(x) and the x-axis from a to b. A definite integral gives the net (signed) area under the graph of f between the limits.

Question 3

A definite integral, unlike an indefinite one:

Accepted Answer

Has a definite numerical value and no constant of integration C. The constant cancels in F(b) − F(a), so a definite integral is a pure number, not a function.

Question 4

The value of \int_a^a f(x)\,dx is:

Accepted Answer

0. Equal limits enclose zero width, so the integral is F(a) − F(a) = 0.

Question 5

Reversing the limits of a definite integral: \int_b^a f(x)\,dx =

Accepted Answer

-\int_a^b f(x)\,dx. Swapping the limits changes the sign of the definite integral.

Question 6

The area under a velocity–time (v–t) graph between two instants gives the:

Accepted Answer

Displacement. Since s = ∫v dt, the area under the v–t curve equals the displacement over that interval.

Question 7

The area under a force–displacement (F–x) graph gives the:

Accepted Answer

Work done by the force. W = ∫F dx, so the area under the F–x curve is the work done.

Question 8

The area under an acceleration–time (a–t) graph gives the:

Accepted Answer

Change in velocity. Since Δv = ∫a dt, the area under the a–t graph is the change in velocity.

Question 9

If a curve lies below the x-axis over [a, b], the definite integral there is:

Accepted Answer

Negative (the signed area is negative). Below the axis f(x) < 0, so the signed area (and the integral) is negative. Pitfall: total geometric area needs |f|.

Question 10

The additivity property of definite integrals states \int_a^b f\,dx =

Accepted Answer

\int_a^c f\,dx + \int_c^b f\,dx. An interval can be split at any interior point c: the integral over [a,b] is the sum of the parts.

Question 11

\int_0^1 x\,dx =

Accepted Answer

\dfrac{1}{2}. [x²/2]₀¹ = 1/2 − 0 = 1/2.

Question 12

\int_0^2 x^2\,dx =

Accepted Answer

\dfrac{8}{3}. [x³/3]₀² = 8/3 − 0 = 8/3.

Question 13

\int_1^3 2x\,dx =

Accepted Answer

8. [x²]₁³ = 9 − 1 = 8.

Question 14

\int_0^{\pi/2} \cos x\,dx =

Accepted Answer

1. [sin x]₀^{π/2} = sin(π/2) − sin0 = 1 − 0 = 1.

Question 15

\int_0^{\pi} \sin x\,dx =

Accepted Answer

2. [−cos x]₀^π = −cos π − (−cos 0) = −(−1) + 1 = 2.

Question 16

\int_0^1 e^x\,dx =

Accepted Answer

e - 1. [eˣ]₀¹ = e¹ − e⁰ = e − 1.

Question 17

\int_1^2 \dfrac{1}{x}\,dx =

Accepted Answer

\ln 2. [ln|x|]₁² = ln2 − ln1 = ln2.

Question 18

\int_0^3 (x^2 + 1)\,dx =

Accepted Answer

12. [x³/3 + x]₀³ = (9 + 3) − 0 = 12.

Question 19

A particle moves with velocity v = 2t (m/s). Its displacement from t = 0 to t = 3 s is:

Accepted Answer

9 m. s = ∫₀³ 2t dt = [t²]₀³ = 9 m. (Also the area of the triangle: ½ × 3 × 6 = 9 m.)

Question 20

A variable force F = 4x (N) acts on a body. The work done as it moves from x = 0 to x = 2 m is:

Accepted Answer

8 J. W = ∫₀² 4x dx = [2x²]₀² = 8 J.

Definite Integrals & Area Under Curve — IISER Basic Maths MCQs with Solutions

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