CBSE Class 12 Mathematics Integrals MCQs Set 05

Answer: (c) 1/3

Explanation:
1. \( e^{2\ln x} = e^{\ln x^2} = x^2 \) for x > 0.
2. \( \int_{0}^{1} x^2\,dx = \left[\frac{x^3}{3}\right]_{0}^{1} = \frac{1}{3} \).

Teacher's Note:
1. Always simplify \( e^{k\ln x} \) to \( x^k \) before integrating; it turns an exponential-looking integrand into a power.
2. A common slip is to write \( e^{2\ln x} = 2x \); the exponent becomes a power, not a multiplier.

Answer: (c) \( \frac{e^2}{2} - e \)

Explanation:
1. The integrand has the form \( e^x[f(x) + f'(x)] \) with \( f(x) = \frac{1}{x} \), since \( f'(x) = -\frac{1}{x^2} \).
2. So \( \int e^x\left(\frac{1}{x} - \frac{1}{x^2}\right)dx = \frac{e^x}{x} \).
3. \( \left[\frac{e^x}{x}\right]_{1}^{2} = \frac{e^2}{2} - e \).

Teacher's Note:
1. Remember \( \int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C \); spotting the pair f and f′ saves integration by parts.
2. Check the sign: the value is positive because \( e^2/2 \approx 3.69 > e \approx 2.72 \).

Answer: (c) \( \log \frac{4}{3} \)

Explanation:
1. \( \frac{1}{x^2 - x} = \frac{1}{x(x - 1)} = \frac{1}{x - 1} - \frac{1}{x} \).
2. \( \int_{2}^{3} = \left[\log\frac{x - 1}{x}\right]_{2}^{3} = \log\frac{2}{3} - \log\frac{1}{2} \).
3. \( = \log\frac{4}{3} \).

Teacher's Note:
1. Partial fractions of \( \frac{1}{x(x-1)} \) follow the quick rule: difference of reciprocals of the factors.
2. Since the integrand is positive on [2, 3], the answer must be positive; this rules out log(2/3) and log(1/4) at once.

Answer: (d) 0

Explanation:
1. Put t = x − 2; the limits 1 to 3 become −1 to 1 and the integrand becomes (t + 1)t(t − 1) = t³ − t.
2. t³ − t is an odd function, so \( \int_{-1}^{1} (t^3 - t)\,dt = 0 \).

Teacher's Note:
1. When a cubic has roots placed symmetrically about the midpoint of the limits, shift to that midpoint and use oddness.
2. Direct expansion also works: x³ − 6x² + 11x − 6 integrates to 0 over [1, 3].

Answer: (a) \( \pi/12 \)

Explanation:
1. \( \int \frac{dx}{1 + x^2} = \tan^{-1} x \).
2. \( \left[\tan^{-1} x\right]_{1}^{\sqrt{3}} = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12} \).

Teacher's Note:
1. Learn the standard values \( \tan^{-1} 1 = \frac{\pi}{4} \) and \( \tan^{-1}\sqrt{3} = \frac{\pi}{3} \) for quick substitution.
2. Do not stop at the upper-limit value \( \frac{\pi}{3} \); the lower limit must be subtracted.

Answer: (c) 1

Explanation:
1. \( \int \frac{1}{x}\,dx = \log x \).
2. \( [\log x]_{1}^{e} = \log e - \log 1 = 1 - 0 = 1 \).

Teacher's Note:
1. log e = 1 and log 1 = 0 are the two values used most often in definite integrals.
2. The integrand is finite on [1, e], so the value cannot be ∞.

Answer: (b) \( \pi/24 \)

Explanation:
1. \( \int_{0}^{2/3} \frac{dx}{4 + 9x^2} = \frac{1}{9}\int_{0}^{2/3} \frac{dx}{x^2 + (2/3)^2} \).
2. \( = \frac{1}{9}\cdot\frac{3}{2}\left[\tan^{-1}\frac{3x}{2}\right]_{0}^{2/3} = \frac{1}{6}\cdot\frac{\pi}{4} \).
3. \( = \frac{\pi}{24} \).

Teacher's Note:
1. Take out the coefficient of x² first so the integrand matches \( \frac{1}{x^2 + a^2} \), then use \( \frac{1}{a}\tan^{-1}\frac{x}{a} \).
2. Forgetting the factor 1/9 gives a value nine times too large, a very common error.

Answer: (c) 8

Explanation:
1. \( 1 + \sin\frac{x}{2} = \left(\sin\frac{x}{4} + \cos\frac{x}{4}\right)^2 \), and this bracket is ≥ 0 for 0 ≤ x ≤ 2π.
2. So \( I = \int_{0}^{2\pi} \left(\sin\frac{x}{4} + \cos\frac{x}{4}\right)dx = \left[-4\cos\frac{x}{4} + 4\sin\frac{x}{4}\right]_{0}^{2\pi} \).
3. \( = (0 + 4) - (-4 + 0) = 8 \).

Teacher's Note:
1. Use \( 1 + \sin\theta = \left(\sin\frac{\theta}{2} + \cos\frac{\theta}{2}\right)^2 \) to remove the square root.
2. Before dropping the root, check that the bracket stays non-negative on the whole interval; otherwise a modulus is needed.

Answer: (c) 2

Explanation:
1. \( \sqrt{1 + \sin 2x} = \sqrt{(\sin x + \cos x)^2} = \sin x + \cos x \) for 0 ≤ x ≤ π/2.
2. So \( I = \int_{0}^{\pi/2} (\sin x + \cos x)\,dx = [-\cos x + \sin x]_{0}^{\pi/2} \).
3. \( = (0 + 1) - (-1 + 0) = 2 \).

Teacher's Note:
1. 1 + sin 2x = (sin x + cos x)² is the key identity whenever √(1 ± sin 2x) appears.
2. sin x + cos x is positive on [0, π/2], so no modulus sign is needed here.

Answer: (b) \( \frac{1}{2}(e^{\pi/2} + 1) \)

Explanation:
1. \( \int e^x \sin x\,dx = \frac{e^x}{2}(\sin x - \cos x) \).
2. At π/2: \( \frac{e^{\pi/2}}{2}(1 - 0) \); at 0: \( \frac{1}{2}(0 - 1) = -\frac{1}{2} \).
3. Value \( = \frac{e^{\pi/2}}{2} + \frac{1}{2} = \frac{1}{2}(e^{\pi/2} + 1) \).

Teacher's Note:
1. Use the formula \( \int e^{ax}\sin bx\,dx = \frac{e^{ax}}{a^2 + b^2}(a\sin bx - b\cos bx) \) instead of repeating parts twice.
2. The integrand is positive on (0, π/2), so option (c), which is negative, can be rejected quickly.

Answer: (c) \( \log \frac{4}{e} \)

Explanation:
1. By parts, \( \int \log x\,dx = x\log x - x \).
2. \( [x\log x - x]_{1}^{2} = (2\log 2 - 2) - (0 - 1) = 2\log 2 - 1 \).
3. \( = \log 4 - \log e = \log\frac{4}{e} \).

Teacher's Note:
1. ∫ log x dx = x log x − x is a result worth memorising.
2. Write the constant 1 as log e to combine the answer into a single logarithm.

Answer: (c) \( \pi/2 \)

Explanation:
1. By the fundamental theorem, \( \int_{0}^{1} \frac{d}{dx}[F(x)]\,dx = F(1) - F(0) \).
2. \( F(1) = \sin^{-1} 1 = \frac{\pi}{2} \) and \( F(0) = \sin^{-1} 0 = 0 \).
3. So the value is \( \frac{\pi}{2} \).

Teacher's Note:
1. Integrating a derivative just returns the function, so evaluate it at the limits without differentiating.
2. For 0 ≤ x ≤ 1, \( \sin^{-1}\frac{2x}{1 + x^2} = 2\tan^{-1} x \), which also gives \( 2\cdot\frac{\pi}{4} = \frac{\pi}{2} \).

Answer: (c) The value of c

Explanation:
1. ax³ and bx are odd functions, so \( \int_{-2}^{2} (ax^3 + bx)\,dx = 0 \).
2. \( \int_{-2}^{2} c\,dx = 4c \).
3. Hence the value is 4c, which depends only on c.

Teacher's Note:
1. Over a symmetric interval [−a, a], odd terms vanish and even terms double: \( \int_{-a}^{a} f = 2\int_{0}^{a} f \) for even f.
2. Split a polynomial into odd and even parts before integrating over symmetric limits.

Answer: (c) 2

Explanation:
1. \( \frac{1}{1 + \sin x} = \frac{1 - \sin x}{\cos^2 x} = \sec^2 x - \sec x\tan x \).
2. Its antiderivative is tan x − sec x \( = \frac{\sin x - 1}{\cos x} = -\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \), continuous on [0, π].
3. Value \( = \left[-\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right]_{0}^{\pi} = -(-1) - (-1) = 2 \).

Teacher's Note:
1. The form \( -\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \) avoids the point x = π/2 where tan x and sec x separately blow up.
2. This is a standard result: ∫₀^π dx/(1 + sin x) = 2.

Answer: (b) 1

Explanation:
1. By parts, \( \int \cos^{-1} x\,dx = x\cos^{-1} x - \sqrt{1 - x^2} \).
2. \( \left[x\cos^{-1} x - \sqrt{1 - x^2}\right]_{0}^{1} = (0 - 0) - (0 - 1) = 1 \).

Teacher's Note:
1. Equivalently, \( \int_{0}^{1} \cos^{-1} x\,dx = \int_{0}^{\pi/2} \cos y\,dy = 1 \) by reading the area along the y-axis.
2. Remember \( \cos^{-1} 1 = 0 \), which makes the upper-limit term vanish.

Answer: (a) \( \frac{\pi}{4} - J \)

Explanation:
1. \( I + J = \int_{0}^{\pi/4} (\sin^2 x + \cos^2 x)\,dx = \int_{0}^{\pi/4} 1\,dx = \frac{\pi}{4} \).
2. Hence \( I = \frac{\pi}{4} - J \).

Teacher's Note:
1. Adding or subtracting companion integrals often removes the hard part; here sin²x + cos²x = 1.
2. I ≠ J here because the interval is [0, π/4], not [0, π/2]; on [0, π/2] they would be equal.

Answer: (a) \( \frac{1}{4} \log \frac{32}{17} \)

Explanation:
1. \( \phi(x) = \frac{1}{x(x^4 + 1)} = \frac{x^3}{x^4(x^4 + 1)} \); put \( t = x^4 \), so \( x^3\,dx = \frac{dt}{4} \).
2. \( \int_{1}^{2} \phi(x)\,dx = \frac{1}{4}\int_{1}^{16} \frac{dt}{t(t + 1)} = \frac{1}{4}\left[\log\frac{t}{t + 1}\right]_{1}^{16} \).
3. \( = \frac{1}{4}\left(\log\frac{16}{17} - \log\frac{1}{2}\right) = \frac{1}{4}\log\frac{32}{17} \).

Teacher's Note:
1. For \( \frac{1}{x(x^n + 1)} \), multiply top and bottom by \( x^{n-1} \) and substitute \( t = x^n \).
2. Remember to change the limits: x = 1, 2 give t = 1, 16.

Answer: (d) \( \frac{\pi}{6} \)

Explanation:
1. \( x - x^2 = \frac{1}{4} - \left(x - \frac{1}{2}\right)^2 \).
2. \( \int \frac{dx}{\sqrt{(1/2)^2 - (x - 1/2)^2}} = \sin^{-1}(2x - 1) \).
3. \( [\sin^{-1}(2x - 1)]_{1/4}^{1/2} = \sin^{-1} 0 - \sin^{-1}\left(-\frac{1}{2}\right) = 0 + \frac{\pi}{6} = \frac{\pi}{6} \).

Teacher's Note:
1. Complete the square under the root to reach the standard form 1/√(a² − u²).
2. Note \( \sin^{-1}(-\theta) = -\sin^{-1}\theta \), which turns the subtraction into an addition.

Answer: (b) \( \frac{4\sqrt{2}}{3} \)

Explanation:
1. Rationalise: \( \frac{1}{\sqrt{1 + x} - \sqrt{x}} = \frac{\sqrt{1 + x} + \sqrt{x}}{(1 + x) - x} = \sqrt{1 + x} + \sqrt{x} \).
2. \( \int_{0}^{1} = \left[\frac{2}{3}(1 + x)^{3/2} + \frac{2}{3}x^{3/2}\right]_{0}^{1} = \frac{2}{3}(2\sqrt{2} + 1) - \frac{2}{3} \).
3. \( = \frac{4\sqrt{2}}{3} \).

Teacher's Note:
1. Rationalising a difference of square roots whose squares differ by a constant gives a simple sum to integrate.
2. Do not forget the lower-limit term (2/3)(1)^(3/2), which cancels the extra 2/3.

Answer: (a) 0

Explanation:
1. \( \int (\sin x + \cos x)\,dx = -\cos x + \sin x \).
2. \( [-\cos x + \sin x]_{0}^{2\pi} = (-1 + 0) - (-1 + 0) = 0 \).

Teacher's Note:
1. Over a full period, sin x and cos x each integrate to zero.
2. This is a signed value; the area enclosed by the curve over [0, 2π] is not zero.

Answer: (a) \( \log(2\sqrt{2}) + \frac{\pi}{12} \)

Explanation:
1. Split: \( \int_{0}^{3} \frac{3x}{x^2 + 9}\,dx + \int_{0}^{3} \frac{dx}{x^2 + 9} \).
2. First part \( = \frac{3}{2}[\log(x^2 + 9)]_{0}^{3} = \frac{3}{2}\log 2 = \log 2\sqrt{2} \).
3. Second part \( = \frac{1}{3}\left[\tan^{-1}\frac{x}{3}\right]_{0}^{3} = \frac{1}{3}\cdot\frac{\pi}{4} = \frac{\pi}{12} \).
4. Total \( = \log(2\sqrt{2}) + \frac{\pi}{12} \).

Teacher's Note:
1. For (linear)/(quadratic), split the numerator into a multiple of the derivative of the denominator plus a constant.
2. Note \( \frac{3}{2}\log 2 = \log 2^{3/2} = \log 2\sqrt{2} \).

Answer: (a) \( 1 - \frac{\pi}{4} \)

Explanation:
1. \( \tan^2 x = \sec^2 x - 1 \).
2. \( \int_{0}^{\pi/4} (\sec^2 x - 1)\,dx = [\tan x - x]_{0}^{\pi/4} = 1 - \frac{\pi}{4} \).

Teacher's Note:
1. Convert tan²x to sec²x − 1 because sec²x has a direct antiderivative.
2. Since the integrand is positive, the result must be positive; \( \frac{\pi}{4} - 1 \) is negative and can be rejected.

Answer: (c) \( \pi/2 \)

Explanation:
1. \( \frac{x + \sin x}{1 + \cos x} = \frac{x}{2\cos^2(x/2)} + \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} = \frac{x}{2}\sec^2\frac{x}{2} + \tan\frac{x}{2} \).
2. This is \( \frac{d}{dx}\left[x\tan\frac{x}{2}\right] \), so the integral is \( \left[x\tan\frac{x}{2}\right]_{0}^{\pi/2} \).
3. \( = \frac{\pi}{2}\tan\frac{\pi}{4} - 0 = \frac{\pi}{2} \).

Teacher's Note:
1. Half-angle formulas 1 + cos x = 2cos²(x/2) and sin x = 2 sin(x/2) cos(x/2) are the key here.
2. The integrand is (x·g′ + g) with g = tan(x/2), i.e. the derivative of x·g; recognising this avoids integration by parts.

Answer: (a) \( \int_{0}^{1} (f(x) + g(x))\,dx = M + N \)

Explanation:
1. Linearity: \( \int_{0}^{1} (f + g)\,dx = \int_{0}^{1} f\,dx + \int_{0}^{1} g\,dx = M + N \).
2. No such rule exists for products, reciprocals or quotients; e.g. f = g = x gives ∫x² = 1/3 but MN = 1/4.
3. So only option (a) is always correct.

Teacher's Note:
1. The definite integral is linear: it splits over sums and lets constants come out, but nothing more.
2. A single counterexample like f(x) = g(x) = x is enough to reject a proposed rule.

Answer: (a) \( -\frac{1}{2} e^{-\pi/2} \)

Explanation:
1. \( \int e^{-x}\sin x\,dx = \frac{e^{-x}}{2}(-\sin x - \cos x) = -\frac{e^{-x}}{2}(\sin x + \cos x) \).
2. At π/2: \( -\frac{e^{-\pi/2}}{2}(1 + 0) = -\frac{1}{2}e^{-\pi/2} \).
3. At −π/4: sin x + cos x = 0, so the term is 0. Value \( = -\frac{1}{2}e^{-\pi/2} \).

Teacher's Note:
1. Use \( \int e^{ax}\sin bx\,dx = \frac{e^{ax}}{a^2 + b^2}(a\sin bx - b\cos bx) \) with a = −1, b = 1.
2. Check where the bracket vanishes: sin x + cos x = 0 at x = −π/4 makes the lower-limit term disappear.

Answer: (a) \( \frac{3}{2} \)

Explanation:
1. Put t = 1 + log x, so \( dt = \frac{dx}{x} \); x = 1 → t = 1 and x = e → t = 2.
2. \( \int_{1}^{2} t\,dt = \left[\frac{t^2}{2}\right]_{1}^{2} = 2 - \frac{1}{2} = \frac{3}{2} \).

Teacher's Note:
1. Whenever log x and 1/x appear together, substitute t = log x (or 1 + log x).
2. Change the limits along with the variable so you need not substitute back.

Answer: (c) \( a = \frac{3}{4}, b = \frac{3}{2} \)

Explanation:
1. By parts: \( \int_{0}^{1} x\log\left(1 + \frac{x}{2}\right)dx = \left[\frac{x^2}{2}\log\frac{2 + x}{2}\right]_{0}^{1} - \int_{0}^{1} \frac{x^2}{2(x + 2)}\,dx \).
2. First term \( = \frac{1}{2}\log\frac{3}{2} \). Also \( \frac{x^2}{x + 2} = x - 2 + \frac{4}{x + 2} \), so \( \int_{0}^{1} \frac{x^2}{x + 2}dx = \frac{1}{2} - 2 + 4\log\frac{3}{2} \).
3. Integral \( = \frac{1}{2}\log\frac{3}{2} + \frac{3}{4} - 2\log\frac{3}{2} = \frac{3}{4} - \frac{3}{2}\log\frac{3}{2} = \frac{3}{4} + \frac{3}{2}\log\frac{2}{3} \).
4. So \( a = \frac{3}{4}, b = \frac{3}{2} \).

Teacher's Note:
1. Take the logarithm as the first function (ILATE) and the algebraic factor x as the second.
2. Use \( \log\frac{3}{2} = -\log\frac{2}{3} \) to match the required form before reading off b.

Answer: (a) \( \frac{1}{2} \log 2 \)

Explanation:
1. \( \frac{1 - \tan x}{1 + \tan x} = \tan\left(\frac{\pi}{4} - x\right) \).
2. \( \int_{0}^{\pi/4} \tan\left(\frac{\pi}{4} - x\right)dx = \left[\log\cos\left(\frac{\pi}{4} - x\right)\right]_{0}^{\pi/4} \).
3. \( = \log 1 - \log\frac{1}{\sqrt{2}} = \frac{1}{2}\log 2 \).

Teacher's Note:
1. Recognise \( \frac{1 - \tan x}{1 + \tan x} = \tan\left(\frac{\pi}{4} - x\right) \); by the a − x property this equals \( \int_{0}^{\pi/4} \tan x\,dx \).
2. ∫ tan u du = −log cos u; be careful with the sign coming from the chain rule.

Answer: (b) \( \pi/4 \)

Explanation:
1. \( x^2 + 2x + 2 = (x + 1)^2 + 1 \).
2. \( \int_{-1}^{0} \frac{dx}{(x + 1)^2 + 1} = [\tan^{-1}(x + 1)]_{-1}^{0} = \tan^{-1} 1 - \tan^{-1} 0 = \frac{\pi}{4} \).

Teacher's Note:
1. Complete the square in the denominator to reach 1/(u² + 1).
2. The integrand is positive, so the negative option −π/4 is impossible.

Answer: (d) \( \frac{1}{36}(\pi^2 + 16) \)

Explanation:
1. \( \int (2 + 3x^2)\cos 3x\,dx = (2 + 3x^2)\frac{\sin 3x}{3} - \int 6x\frac{\sin 3x}{3}dx = (2 + 3x^2)\frac{\sin 3x}{3} + \frac{2x\cos 3x}{3} - \frac{2\sin 3x}{9} \).
2. At π/6: sin 3x = 1, cos 3x = 0, giving \( \frac{1}{3}\left(2 + \frac{\pi^2}{12}\right) - \frac{2}{9} = \frac{4}{9} + \frac{\pi^2}{36} \); at 0 everything is 0.
3. Value \( = \frac{16 + \pi^2}{36} = \frac{1}{36}(\pi^2 + 16) \).

Teacher's Note:
1. Apply parts twice (or the tabular method), differentiating the polynomial until it becomes zero.
2. Since x² appears in the integrand, the answer must contain π²; this quickly eliminates options (a) and (b).

Answer: (d) All of the above

Explanation:
1. \( \int_{\pi/2}^{\alpha} \sin x\,dx = [-\cos x]_{\pi/2}^{\alpha} = -\cos\alpha \).
2. So −cos α = sin 2α = 2 sin α cos α, i.e. \( \cos\alpha(1 + 2\sin\alpha) = 0 \).
3. cos α = 0 gives α = π/2, 3π/2; sin α = −1/2 gives α = 7π/6, 11π/6.
4. Hence π/2, 3π/2 and 7π/6 all satisfy it: all of the above.

Teacher's Note:
1. Move everything to one side and factor; never cancel cos α, since that loses the roots where cos α = 0.
2. Check each option by direct substitution when the options are specific values.

Answer: (d) \( 4\sqrt{2} - 2 \)

Explanation:
1. Let F(x) = sin x + cos x, an antiderivative of cos x − sin x. First integral \( = F(\pi/4) - F(0) = \sqrt{2} - 1 \).
2. Second \( = -[F(5\pi/4) - F(\pi/4)] = -(-\sqrt{2} - \sqrt{2}) = 2\sqrt{2} \).
3. Third \( = F(\pi/4) - F(2\pi) = \sqrt{2} - 1 \).
4. Sum \( = (\sqrt{2} - 1) + 2\sqrt{2} + (\sqrt{2} - 1) = 4\sqrt{2} - 2 \).

Teacher's Note:
1. Notice the third integral runs from 2π down to π/4; just apply F(upper) − F(lower) as written.
2. Using one antiderivative for all pieces keeps the arithmetic short and reduces sign errors.

Answer: (b) −2 ≤ a ≤ 4

Explanation:
1. \( \int_{0}^{a} x\,dx = \frac{a^2}{2} \).
2. \( \frac{a^2}{2} \le a + 4 \Rightarrow a^2 - 2a - 8 \le 0 \Rightarrow (a - 4)(a + 2) \le 0 \).
3. Hence −2 ≤ a ≤ 4.

Teacher's Note:
1. Evaluate the integral first; the problem then reduces to an ordinary quadratic inequality.
2. For (a − p)(a − q) ≤ 0 with p < q, the solution lies between the roots.

Answer: (a) \( \pi^2/8 \)

Explanation:
1. For 0 ≤ x < π/2, 0 ≤ sin x < 1, so [sin x] = 0 (it equals 1 only at the single point x = π/2).
2. So the integral \( = \int_{0}^{\pi/2} x\,dx = \frac{1}{2}\left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{8} \).

Teacher's Note:
1. A change of value at a single point does not affect a definite integral.
2. For greatest-integer integrands, first find the sub-intervals where the bracket is constant.

Answer: (a) \( \frac{1}{6}(3\pi - 4) \)

Explanation:
1. Divide: \( \frac{x^4 + 1}{x^2 + 1} = x^2 - 1 + \frac{2}{x^2 + 1} \).
2. \( \int_{0}^{1} = \left[\frac{x^3}{3} - x + 2\tan^{-1} x\right]_{0}^{1} = \frac{1}{3} - 1 + \frac{\pi}{2} \).
3. \( = \frac{\pi}{2} - \frac{2}{3} = \frac{1}{6}(3\pi - 4) \).

Teacher's Note:
1. When the numerator's degree is not less than the denominator's, divide first.
2. A quick check: the integrand lies between 1/2 and 1 on [0, 1], and (3π − 4)/6 ≈ 0.90 fits.

Answer: (c) −5

Explanation:
1. \( \int_{2}^{4} (3 - f)\,dx = 6 - \int_{2}^{4} f\,dx = 7 \), so \( \int_{2}^{4} f\,dx = -1 \).
2. \( \int_{-1}^{2} f\,dx = \int_{-1}^{4} f\,dx - \int_{2}^{4} f\,dx = 4 - (-1) = 5 \).
3. \( \int_{2}^{-1} f\,dx = -\int_{-1}^{2} f\,dx = -5 \).

Teacher's Note:
1. Use \( \int_{a}^{b} = \int_{a}^{c} + \int_{c}^{b} \) to split intervals, and \( \int_{b}^{a} = -\int_{a}^{b} \) to reverse limits.
2. The constant 3 integrates to 3 × (4 − 2) = 6; forgetting the interval length is a common slip.

Answer: (b) \( 2 + \log_e\left(\frac{15}{7}\right) \)

Explanation:
1. \( \frac{x^2}{x^2 - 4} = 1 + \frac{4}{x^2 - 4} = 1 + \frac{1}{x - 2} - \frac{1}{x + 2} \).
2. \( \int_{3}^{5} = \left[x + \log\frac{x - 2}{x + 2}\right]_{3}^{5} = 2 + \log\frac{3}{7} - \log\frac{1}{5} \).
3. \( = 2 + \log\frac{15}{7} \).

Teacher's Note:
1. Use \( \int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\log\left|\frac{x - a}{x + a}\right| \); here 4 × (1/4) gives coefficient 1.
2. The integrand exceeds 1 on [3, 5], so the value must exceed 2; this rules out option (a).

Answer: (a) 0

Explanation:
1. Expand: \( \int_{0}^{1} (x - a)^2 f\,dx = \int_{0}^{1} x^2 f\,dx - 2a\int_{0}^{1} x f\,dx + a^2\int_{0}^{1} f\,dx \).
2. \( = a^2 - 2a\cdot a + a^2\cdot 1 = 0 \).

Teacher's Note:
1. Expand the square and use linearity; a is a constant and comes outside the integral.
2. In statistics language, this is a variance computed to be zero from the given moments.

Answer: (b) \( \frac{e}{4} - 1 \)

Explanation:
1. Write \( \frac{x - 1}{(x + 1)^3} = \frac{(x + 1) - 2}{(x + 1)^3} = \frac{1}{(x + 1)^2} - \frac{2}{(x + 1)^3} \).
2. With \( f(x) = \frac{1}{(x + 1)^2} \), \( f'(x) = -\frac{2}{(x + 1)^3} \), so the integrand is \( e^x[f(x) + f'(x)] \).
3. Value \( = \left[\frac{e^x}{(x + 1)^2}\right]_{0}^{1} = \frac{e}{4} - 1 \).

Teacher's Note:
1. Rewrite the numerator in terms of the denominator's base (x + 1) to reveal the f + f′ pattern.
2. The integrand is negative on [0, 1), so the answer must be negative; e/4 − 1 ≈ −0.32 fits.

Answer: (c) \( \frac{\pi}{4} - \frac{1}{2} \log 2 \)

Explanation:
1. Let I be the given integral and J the one with sin x in the numerator. By x → π/2 − x, I = J.
2. \( I + J = \int_{0}^{\pi/2} \frac{\sin x + \cos x}{1 + \sin x + \cos x}dx = \int_{0}^{\pi/2} \left(1 - \frac{1}{1 + \sin x + \cos x}\right)dx \).
3. With t = tan(x/2), \( \int_{0}^{\pi/2} \frac{dx}{1 + \sin x + \cos x} = \int_{0}^{1} \frac{dt}{1 + t} = \log 2 \).
4. So 2I = π/2 − log 2, giving \( I = \frac{\pi}{4} - \frac{1}{2}\log 2 \).

Teacher's Note:
1. Pairing with the complementary integral (replace x by π/2 − x) is the standard trick for such sin/cos integrands.
2. Remember \( \int_{0}^{\pi/2} \frac{dx}{1 + \sin x + \cos x} = \log 2 \) as a useful result.

Answer: (c) \( \frac{1}{2} e^{\pi/4} \log 2 \)

Explanation:
1. With \( f(x) = \log\sin x \), \( f'(x) = \cot x \), so the integrand is \( e^x[f(x) + f'(x)] \).
2. Value \( = [e^x\log\sin x]_{\pi/4}^{\pi/2} = e^{\pi/2}\log 1 - e^{\pi/4}\log\frac{1}{\sqrt{2}} \).
3. \( = 0 + \frac{1}{2}e^{\pi/4}\log 2 = \frac{1}{2}e^{\pi/4}\log 2 \).

Teacher's Note:
1. Spot \( e^x[f + f'] \) by checking whether one term in the bracket is the derivative of the other.
2. \( \log\frac{1}{\sqrt{2}} = -\frac{1}{2}\log 2 \); the minus sign then cancels with the subtraction.

Answer: (d) \( \frac{1}{\pi} \)

Explanation:
1. For 0 ≤ x ≤ 1/2, 0 ≤ πx ≤ π/2, so sin πx ≥ 0 and |sin πx| = sin πx.
2. \( \int_{0}^{1/2} \sin\pi x\,dx = \left[-\frac{\cos\pi x}{\pi}\right]_{0}^{1/2} = -\frac{0}{\pi} + \frac{1}{\pi} = \frac{1}{\pi} \).

Teacher's Note:
1. Before removing a modulus, check the sign of the expression over the whole interval.
2. Do not forget the factor 1/π from the chain rule when integrating sin πx.

Answer: (b) \( \frac{16}{5} \)

Explanation:
1. Put t = 1 − x: \( \int_{0}^{1} \frac{x}{(1 - x)^{3/4}}dx = \int_{0}^{1} \frac{1 - t}{t^{3/4}}dt = \int_{0}^{1} (t^{-3/4} - t^{1/4})\,dt \).
2. \( = \left[4t^{1/4} - \frac{4}{5}t^{5/4}\right]_{0}^{1} = 4 - \frac{4}{5} = \frac{16}{5} \).

Teacher's Note:
1. Substituting for the expression inside the awkward power turns the integrand into simple powers of t.
2. The integrand is positive on (0, 1), so the negative option is impossible.