Integral Calculus & Differential Equations — NDA Mathematics — Practice

Question 1

The value of ∫(5x⁴ − 3x² + 2) dx is:
(a) x⁵ − x³ + 2x + C  (b) 5x⁵ − 3x³ + 2x + C  (c) x⁵ − x³ + 2x  (d) x⁵ + x³ + 2x + C

Accepted Answer

Integrate term by term using ∫xⁿdx = xⁿ⁺¹/(n+1): ∫5x⁴dx = x⁵, ∫−3x²dx = −x³, ∫2dx = 2x.

So the integral is x⁵ − x³ + 2x + C.

Answer: (a) x⁵ − x³ + 2x + C. Trap (b) keeps the original coefficient without dividing by the new power (n+1) — a rushed application of the power rule. Trap (c) is the correct antiderivative with the integration constant dropped entirely — always add +C for an indefinite integral. Trap (d) has a sign error, treating −3x² as if it integrated to +x³.

Question 2

∫ dx/(9+x²) equals:
(a) (1/3)tan⁻¹(x/3) + C  (b) tan⁻¹(x/3) + C  (c) (1/3)tan⁻¹(x/9) + C  (d) (1/9)tan⁻¹(x/3) + C

Accepted Answer

Match to the standard form ∫dx/(a²+x²) = (1/a)tan⁻¹(x/a) + C with a² = 9, so a = 3.

∫dx/(9+x²) = (1/3)tan⁻¹(x/3) + C.

Answer: (a). Trap (b) drops the 1/a multiplier out front entirely. Trap (c) correctly keeps 1/3 out front but wrongly uses a = 9 (instead of a = 3, forgetting to take the square root) inside the tan⁻¹ argument. Trap (d) makes the mirror-image mistake — uses a² = 9 as the outer multiplier (1/9) instead of a = 3.

Question 3

∫ x² e^(x³) dx equals:
(a) (1/3)e^(x³) + C  (b) e^(x³) + C  (c) (1/3)x³e^(x³) + C  (d) 3e^(x³) + C

Accepted Answer

Let u = x³, so du = 3x²dx, i.e. x²dx = du/3.

∫x²e^(x³)dx = ∫e^u (du/3) = (1/3)e^u + C = (1/3)e^(x³) + C.

Answer: (a). Trap (b) is the single most common substitution error — recognising the form correctly but dropping the 1/3 that comes from du = 3x²dx. Trap (c) confuses this with a product-style answer that doesn't actually differentiate back to the original integrand. Trap (d) multiplies by 3 instead of dividing — inverting the constant instead of correctly isolating x²dx from du.

Question 4

∫ x ln x dx equals:
(a) (x²/2)ln x − x²/4 + C  (b) (x²/2)ln x − x²/2 + C  (c) x ln x − x + C  (d) (x²/2)ln x + x²/4 + C

Accepted Answer

By ILATE, ln x (Logarithmic) is chosen as u before x (Algebraic): u = ln x, dv = x dx ⇒ du = dx/x, v = x²/2.

∫x ln x dx = (x²/2)ln x − ∫(x²/2)(1/x)dx = (x²/2)ln x − ∫(x/2)dx = (x²/2)ln x − x²/4 + C.

Answer: (a). Trap (b) makes an arithmetic slip integrating x/2, writing ∫(x/2)dx = x²/2 instead of x²/4. Trap (c) wrongly applies the plain ∫ln x dx = x ln x − x formula, ignoring the extra factor of x in the original integrand. Trap (d) has a sign error, adding instead of subtracting the second term of the by-parts formula.

Question 5

∫ dx / [(x−1)(x+2)] equals:
(a) (1/3)ln|(x−1)/(x+2)| + C  (b) (1/3)ln|(x+2)/(x−1)| + C  (c) ln|(x−1)/(x+2)| + C  (d) (1/3)[ln|x−1| + ln|x+2|] + C

Accepted Answer

Write 1/[(x−1)(x+2)] = A/(x−1) + B/(x+2). Clearing denominators: 1 = A(x+2) + B(x−1).

At x=1: 1 = 3A ⇒ A = 1/3. At x=−2: 1 = −3B ⇒ B = −1/3.

So the integral is (1/3)ln|x−1| − (1/3)ln|x+2| + C = (1/3)ln|(x−1)/(x+2)| + C.

Answer: (a). Trap (b) has the ratio inverted — the sign error that comes from plugging a root into the wrong bracket and getting B = +1/3 instead of −1/3, which flips which log ends up on top. Trap (c) is the correct combined log with the 1/3 (from A and B) dropped entirely. Trap (d) uses B = +1/3 (adding the two logs instead of subtracting) — exactly the 'sign slip in partial fractions' trap of misreading which constant is negative.

Question 6

The value of ∫₋₃³ (x⁵ − 4x³ + 2x) dx is:
(a) 0  (b) 99  (c) −99  (d) 49.5

Accepted Answer

Every term (x⁵, x³, x) is an odd power of x, so f(x) = x⁵ − 4x³ + 2x is an odd function: f(−x) = −f(x).

By property P6, ∫₋ₐᵃ f(x)dx = 0 for any odd f — no integration needed.

Answer: (a) 0. Trap (b) 99 comes from wrongly treating the function as even and applying ∫₋ₐᵃ = 2∫₀ᵃ instead: 2∫₀³(x⁵−4x³+2x)dx = 2(49.5) = 99. Trap (c) −99 is the same wrong-even-property error with an extra sign slip. Trap (d) 49.5 is ∫₀³(x⁵−4x³+2x)dx alone — evaluating only half the interval and skipping the odd/even check altogether.

Question 7

∫₀^(π/2) [sin³x / (sin³x + cos³x)] dx equals:
(a) π/4  (b) π/2  (c) 0  (d) π

Accepted Answer

Let I = ∫₀^(π/2) sin³x/(sin³x+cos³x) dx. Apply P4 (x → π/2 − x): sin(π/2−x)=cos x and cos(π/2−x)=sin x, so

I = ∫₀^(π/2) cos³x/(cos³x+sin³x) dx = J.

Adding: I + J = ∫₀^(π/2) 1 dx = π/2. Since I = J, 2I = π/2, so I = π/4.

Answer: (a) π/4. Trap (b) π/2 stops at I+J = π/2 and forgets the final step of halving once I=J is established. Trap (c) 0 wrongly applies the odd-function shortcut (P6) to an integrand that is neither odd nor defined on a symmetric interval about 0. Trap (d) π comes from doubling π/2 instead of halving it — the reverse of the correct final step.

Question 8

The area bounded by y = x², the x-axis, x = 1 and x = 3 is:
(a) 26/3 sq units  (b) 9 sq units  (c) −26/3 sq units  (d) 26 sq units

Accepted Answer

Area = ∫₁³ x² dx = [x³/3]₁³ = 27/3 − 1/3 = 26/3.

Answer: (a) 26/3 sq units. Trap (b) 9 evaluates only the antiderivative at the upper limit (x³/3 at x=3) and forgets to subtract its value at the lower limit. Trap (c) −26/3 comes from swapping the limits (F(1) − F(3) instead of F(3) − F(1)) — a sign error, since an area can never actually be negative. Trap (d) 26 forgets to divide by 3 when integrating x², using x³ instead of x³/3 as the antiderivative.

Question 9

The order and degree of the differential equation d²y/dx² = ∛(1 + dy/dx) are respectively:
(a) 2 and 3  (b) 2 and 1  (c) 3 and 2  (d) 2 and 1/3

Accepted Answer

The equation has a derivative under a cube root, so it isn't yet a polynomial in derivatives — cube both sides first to clear the radical: (d²y/dx²)³ = 1 + dy/dx.

Now it's polynomial: the highest-order derivative is d²y/dx² (order 2), and it appears raised to the power 3 (degree 3).

Answer: (a) 2 and 3. Trap (b) 2 and 1 reads the degree straight off the un-rationalised equation — d²y/dx² appears to the 'power 1' before clearing the radical, exactly the 'reading degree before rationalising' trap. Trap (c) swaps the order and degree values. Trap (d) uses the radical's fractional exponent (1/3) directly as the degree instead of cubing to clear it first — degree must always be a positive integer.

Question 10

The differential equation representing the family of curves y = Ae^(2x), where A is an arbitrary constant, is:
(a) dy/dx = 2y  (b) dy/dx = y/2  (c) dy/dx = Ay  (d) d²y/dx² = 2y

Accepted Answer

Differentiate: dy/dx = 2Ae^(2x). Since Ae^(2x) = y, this is dy/dx = 2y — the constant A is fully eliminated.

Answer: (a) dy/dx = 2y. Trap (b) y/2 inverts the constant, dividing instead of multiplying by 2 when substituting back. Trap (c) dy/dx = Ay still contains the arbitrary constant A — the elimination is incomplete, which defeats the entire purpose of 'forming' the DE (a formed DE must be free of the original constants). Trap (d) unnecessarily differentiates a second time even though only one constant needs eliminating, giving the wrong order entirely.

Question 11

If dy/dx = 2xy and y = 3 when x = 0, then y equals:
(a) 3e^(x²)  (b) 3e^(2x)  (c) x² + 3  (d) e^(x²) + 3

Accepted Answer

Separate: dy/y = 2x dx. Integrating both sides: ln|y| = x² + C.

At x=0, y=3: ln 3 = C, so ln y = x² + ln 3, i.e. y = e^(x²+ln3) = 3e^(x²).

Answer: (a) 3e^(x²). Trap (b) 3e^(2x) pattern-matches to the growth formula N=N₀e^(kt) without actually integrating ∫2x dx = x² — it uses the coefficient 2 directly in the exponent instead of the true antiderivative x². Trap (c) x²+3 forgets to exponentiate the log equation back to y — it stops at the ln-stage and treats the constant additively. Trap (d) e^(x²)+3 makes the constant of integration additive instead of correctly folding it in multiplicatively via e^(ln3) = 3.

Question 12

A radioactive substance decays according to dN/dt = −kN. If its half-life is 20 years, the value of the decay constant k is:
(a) (ln 2)/20 per year  (b) 1/10 per year  (c) 20/ln 2 per year  (d) ln 40 per year

Accepted Answer

Separating and integrating dN/N=−kdt gives N=N₀e^(−kt). At the half-life t=T, N=N₀/2:

N₀/2 = N₀e^(−kT) ⇒ ln(1/2) = −kT ⇒ k = (ln 2)/T.

With T = 20 years: k = (ln 2)/20 per year.

Answer: (a). Trap (b) 1/10 drops the 'ln' entirely, using k=2/20=1/10 instead of (ln2)/20. Trap (c) 20/ln2 inverts the correct fraction. Trap (d) ln 40 wrongly merges the two into a single logarithm, computing ln(2×20) instead of the correct (ln2)/20.

Practice — Integral Calculus & Differential Equations — NDA Mathematics

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