Learn › JEE Main · Basic Maths ›
Integration › Substitution & Physics Applications
Substitution & Physics Applications — JEE Main Basic Maths MCQs with Solutions
Free JEE Main Basic Maths Substitution & Physics Applications MCQs with step-by-step solutions (25 questions). Part of Integration. Practise online on Prepizo — no login needed.
▶ Practise Substitution & Physics Applications online (free)
Questions with solutions
Q1 — Substitution & Physics Applications · easy · theory
For a linear inner function, $\int f(ax+b)\,dx =$
A. $F(ax+b) + C$
B. $\dfrac{1}{a}F(ax+b) + C$, where F is an antiderivative of f ✓ Correct
C. $a\,F(ax+b) + C$
D. $\dfrac{1}{a+b}F(ax+b) + C$
Solution: Substituting u = ax + b (du = a dx) introduces a factor 1/a: ∫f(ax+b)dx = (1/a)F(ax+b) + C.
Q2 — Substitution & Physics Applications · easy · theory
In u-substitution, if we set u = g(x), then du equals:
A. $dx$
B. $\dfrac{1}{g'(x)}\,dx$
C. $g(x)\,dx$
D. $g'(x)\,dx$ ✓ Correct
Solution: du = g′(x) dx. The substitution succeeds when g′(x) (up to a constant) already appears in the integrand.
Q3 — Substitution & Physics Applications · easy · theory
Displacement is obtained from velocity by the relation:
A. $s = \dfrac{dv}{dt}$
B. $s = v\,t$ always
C. $s = \int v\,dt$ ✓ Correct
D. $s = \int a\,dt$
Solution: Since v = ds/dt, integrating velocity over time gives displacement: s = ∫v dt.
Q4 — Substitution & Physics Applications · easy · theory
Velocity is obtained from acceleration by:
A. $v = \int a\,dt$ ✓ Correct
B. $v = \int s\,dt$
C. $v = \dfrac{da}{dt}$
D. $v = a t^2$
Solution: Since a = dv/dt, integrating acceleration over time gives velocity: v = ∫a dt (+ initial velocity).
Q5 — Substitution & Physics Applications · easy · theory
The work done by a variable force F(x) over a displacement is:
A. $W = \int F\,dx$ ✓ Correct
B. $W = \int F\,dt$
C. $W = \dfrac{dF}{dx}$
D. $W = F\,x$ always
Solution: For a position-dependent force, W = ∫F dx. (W = Fx only holds for a constant force.)
Q6 — Substitution & Physics Applications · easy · theory
The electric charge that has flowed is related to the current I(t) by:
A. $q = \dfrac{dI}{dt}$
B. $q = \int I\,dx$
C. $q = I t$ always
D. $q = \int I\,dt$ ✓ Correct
Solution: Since I = dq/dt, integrating current over time gives charge: q = ∫I dt.
Q7 — Substitution & Physics Applications · medium · theory
The mass of a rod with a position-dependent linear density λ(x) is:
A. $m = \dfrac{d\lambda}{dx}$
B. $m = \lambda x$ always
C. $m = \int \lambda\,dt$
D. $m = \int \lambda\,dx$ ✓ Correct
Solution: A small element dm = λ dx; integrating over the length gives m = ∫λ dx.
Q8 — Substitution & Physics Applications · medium · theory
The integral $\int \dfrac{g'(x)}{g(x)}\,dx$ equals:
A. $\ln|g(x)| + C$ ✓ Correct
B. $\dfrac{[g(x)]^2}{2} + C$
C. $\dfrac{1}{g(x)} + C$
D. $g(x)\ln g(x) + C$
Solution: With u = g(x), du = g′(x)dx, the integral becomes ∫du/u = ln|u| + C = ln|g(x)| + C — a key substitution pattern.
Q9 — Substitution & Physics Applications · medium · theory
u-substitution is the integration counterpart of which differentiation rule?
A. The quotient rule
B. The product rule
C. The power rule only
D. The chain rule ✓ Correct
Solution: Substitution reverses the chain rule: it "undoes" the inner-function derivative introduced by chaining.
Q10 — Substitution & Physics Applications · easy · theory
The impulse delivered by a time-varying force F(t) over an interval is:
A. $J = \int F\,dt$ ✓ Correct
B. $J = \dfrac{dF}{dt}$
C. $J = \int F\,dx$
D. $J = F/t$
Solution: Impulse J = ∫F dt = change in momentum (impulse–momentum theorem).
Q11 — Substitution & Physics Applications · easy · numerical
$\int (2x+3)^4\,dx =$
A. $\dfrac{(2x+3)^5}{5} + C$
B. $\dfrac{(2x+3)^5}{2} + C$
C. $4(2x+3)^3 + C$
D. $\dfrac{(2x+3)^5}{10} + C$ ✓ Correct
Solution: u = 2x+3, du = 2dx: ∫u⁴(du/2) = u⁵/10 + C = (2x+3)⁵/10 + C.
Q12 — Substitution & Physics Applications · easy · numerical
$\int \cos(3x+1)\,dx =$
A. $3\sin(3x+1) + C$
B. $-\dfrac{\sin(3x+1)}{3} + C$
C. $\dfrac{\sin(3x+1)}{3} + C$ ✓ Correct
D. $\sin(3x+1) + C$
Solution: ∫cos(3x+1)dx = (1/3)sin(3x+1) + C (linear substitution, factor 1/3).
Q13 — Substitution & Physics Applications · easy · numerical
$\int e^{5x-2}\,dx =$
A. $\dfrac{e^{5x-2}}{3} + C$
B. $5e^{5x-2} + C$
C. $\dfrac{e^{5x-2}}{5} + C$ ✓ Correct
D. $e^{5x-2} + C$
Solution: ∫e^{5x−2}dx = e^{5x−2}/5 + C.
Q14 — Substitution & Physics Applications · medium · numerical
$\int 2x(x^2+1)^3\,dx =$
A. $(x^2+1)^4 + C$
B. $\dfrac{(x^2+1)^4}{4} + C$ ✓ Correct
C. $\dfrac{(x^2+1)^3}{3} + C$
D. $\dfrac{(x^2+1)^4}{8} + C$
Solution: u = x²+1, du = 2x dx: ∫u³du = u⁴/4 = (x²+1)⁴/4 + C. Trap: the 2x is exactly du — no extra constant needed.
Q15 — Substitution & Physics Applications · medium · numerical
$\int \dfrac{2x}{x^2+1}\,dx =$
A. $2\ln(x^2+1) + C$
B. $\dfrac{1}{x^2+1} + C$
C. $\ln(2x) + C$
D. $\ln(x^2+1) + C$ ✓ Correct
Solution: The numerator is the derivative of the denominator: ∫(g′/g)dx = ln|g| = ln(x²+1) + C.
Q16 — Substitution & Physics Applications · medium · numerical
$\int \sin x\cos x\,dx =$
A. $\dfrac{\sin^2 x}{2} + C$ ✓ Correct
B. $\sin^2 x + C$
C. $\dfrac{\cos^2 x}{2} + C$
D. $-\dfrac{\cos^2 x}{2} + C$ (also valid up to a constant, but pick the sin form)
Solution: u = sin x, du = cos x dx: ∫u du = sin²x/2 + C.
Q17 — Substitution & Physics Applications · medium · numerical
A particle has velocity v = 3t² (m/s). Its displacement from t = 0 to t = 2 s is:
A. 4 m
B. 12 m
C. 8 m ✓ Correct
D. 6 m
Solution: s = ∫₀² 3t² dt = [t³]₀² = 8 m.
Q18 — Substitution & Physics Applications · medium · numerical
A body starts from rest with acceleration a = 6t (m/s²). Its velocity at t = 2 s is:
A. 24 m/s
B. 18 m/s
C. 6 m/s
D. 12 m/s ✓ Correct
Solution: v = ∫₀² 6t dt = [3t²]₀² = 12 m/s.
Q19 — Substitution & Physics Applications · medium · numerical
A variable force F = 2x² (N) acts on a body. The work done from x = 0 to x = 3 m is:
A. 9 J
B. 18 J ✓ Correct
C. 27 J
D. 54 J
Solution: W = ∫₀³ 2x² dx = [2x³/3]₀³ = 2(27)/3 = 18 J.
Q20 — Substitution & Physics Applications · medium · numerical
A current I = 4t (A) flows in a wire. The charge that passes from t = 0 to t = 3 s is:
A. 18 C ✓ Correct
B. 12 C
C. 9 C
D. 36 C
Solution: q = ∫₀³ 4t dt = [2t²]₀³ = 18 C.
Q21 — Substitution & Physics Applications · medium · numerical
A rod of length 5 m has linear density λ = 2x (kg/m), where x is measured from one end. Its total mass is:
A. 10 kg
B. 50 kg
C. 25 kg ✓ Correct
D. 12.5 kg
Solution: m = ∫₀⁵ 2x dx = [x²]₀⁵ = 25 kg.
Q22 — Substitution & Physics Applications · medium · numerical
$\int (1-2x)^5\,dx =$
A. $\dfrac{(1-2x)^6}{6} + C$
B. $-\dfrac{(1-2x)^6}{12} + C$ ✓ Correct
C. $\dfrac{(1-2x)^6}{12} + C$
D. $-\dfrac{(1-2x)^6}{6} + C$
Solution: u = 1−2x, du = −2dx: ∫u⁵(du/−2) = −u⁶/12 + C = −(1−2x)⁶/12 + C.
Q23 — Substitution & Physics Applications · medium · numerical
A particle has velocity v = 4t − t² (m/s). Its displacement from t = 0 to t = 4 s is:
A. $\dfrac{64}{3}$ m
B. 32 m
C. 16 m
D. $\dfrac{32}{3}$ m ≈ 10.7 m ✓ Correct
Solution: s = ∫₀⁴ (4t − t²) dt = [2t² − t³/3]₀⁴ = 32 − 64/3 = 32/3 ≈ 10.7 m.
Q24 — Substitution & Physics Applications · medium · numerical
$\int 2x\,e^{x^2}\,dx =$
A. $x^2 e^{x^2} + C$
B. $2e^{x^2} + C$
C. $\dfrac{e^{x^2}}{2} + C$
D. $e^{x^2} + C$ ✓ Correct
Solution: u = x², du = 2x dx: ∫eᵘ du = eᵘ + C = e^{x²} + C.
Q25 — Substitution & Physics Applications · medium · numerical
$\int x\sqrt{x^2+1}\,dx =$
A. $\dfrac{(x^2+1)^{1/2}}{2} + C$
B. $(x^2+1)^{3/2} + C$
C. $\dfrac{(x^2+1)^{3/2}}{3} + C$ ✓ Correct
D. $\dfrac{(x^2+1)^{3/2}}{2} + C$
Solution: u = x²+1, du = 2x dx: ∫√u (du/2) = (1/2)(u^{3/2}/(3/2)) = u^{3/2}/3 = (x²+1)^{3/2}/3 + C.