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EAPCET Mathematics · Intermediate second year · Paper IIB

Definite Integrals

Definite Integrals evaluates integrals between limits, uses their properties to shortcut hard integrals, and applies them to areas.

Subject
Mathematics
Taught in
Intermediate second year · Paper IIB
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • ∫abf(x) dx=F(b)−F(a)\displaystyle\int_{a}^{b} f(x)\,dx = F(b) - F(a)
  • ∫abf(x) dx=∫abf(a+b−x) dx\displaystyle\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a + b - x)\,dx
  • ∫−aaf(x) dx=2∫0af(x) dx\displaystyle\int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx for even ff, and 0 for odd ff
  • ∫0π/2sin⁡nx dx=∫0π/2cos⁡nx dx\displaystyle\int_{0}^{\pi/2} \sin^{n}x\,dx = \int_{0}^{\pi/2} \cos^{n}x\,dx
  • Area between curves =∫∣f(x)−g(x)∣ dx= \displaystyle\int |f(x) - g(x)|\,dx
  • Wallis: ∫0π/2sin⁡nx dx=(n−1)(n−3)⋯n(n−2)⋯×(π2 if n is even,1 if n is odd)\displaystyle\int_{0}^{\pi/2} \sin^{n}x\,dx = \dfrac{(n - 1)(n - 3)\cdots}{n(n - 2)\cdots} \times \left(\tfrac{\pi}{2}\text{ if } n \text{ is even}, 1 \text{ if } n \text{ is odd}\right)
  • For ff with period TT: ∫0nTf(x) dx=n∫0Tf(x) dx\displaystyle\int_{0}^{nT} f(x)\,dx = n\int_{0}^{T} f(x)\,dx and ∫aa+nTf(x) dx=n∫0Tf(x) dx\displaystyle\int_{a}^{a + nT} f(x)\,dx = n\int_{0}^{T} f(x)\,dx
  • Leibniz rule: ddx∫h(x)g(x)f(t) dt=f(g(x)) g′(x)−f(h(x)) h′(x)\dfrac{d}{dx}\displaystyle\int_{h(x)}^{g(x)} f(t)\,dt = f(g(x))\,g'(x) - f(h(x))\,h'(x)
  • Limit of a sum: lim⁡n→∞b−an∑r=0n−1f(a+r(b−a)n)=∫abf(x) dx\displaystyle\lim_{n \to \infty} \dfrac{b - a}{n}\sum_{r=0}^{n-1} f\left(a + \dfrac{r(b - a)}{n}\right) = \int_{a}^{b} f(x)\,dx; in particular lim⁡n→∞1n∑f(rn)=∫01f(x) dx\displaystyle\lim_{n \to \infty} \dfrac{1}{n}\sum f\left(\dfrac{r}{n}\right) = \int_{0}^{1} f(x)\,dx
  • For strictly monotonic ff: ∫abf(x) dx+∫f(a)f(b)f−1(x) dx=b f(b)−a f(a)\displaystyle\int_{a}^{b} f(x)\,dx + \int_{f(a)}^{f(b)} f^{-1}(x)\,dx = b\,f(b) - a\,f(a)
  • If ff is increasing on [a,b][a, b]: (b−a)f(a)≤∫abf(x) dx≤(b−a)f(b)(b - a)f(a) \le \displaystyle\int_{a}^{b} f(x)\,dx \le (b - a)f(b)
  • Area between the parabolas y2=4axy^{2} = 4ax and x2=4ayx^{2} = 4ay is 16a23\dfrac{16a^{2}}{3}
  • Area between y=k(x−α)(β−x)y = k(x - \alpha)(\beta - x) and the xx-axis is ∣k∣(β−α)36\dfrac{|k|(\beta - \alpha)^{3}}{6}

What this chapter covers

  • Evaluation of definite integrals
  • Properties of definite integrals
  • Reduction formulae
  • Area under and between curves
  • Definite integral as the limit of a sum

How it is usually asked

  • Evaluate using the a + b − x property
  • Find the area between curves
  • Evaluate a limit of a sum as an integral

Understanding Definite Integrals

A definite integral ∫ₐᵇ f(x) dx is a number. By the fundamental theorem of calculus it equals F(b) − F(a), where F is any antiderivative of f, so the constant of integration cancels. Geometrically it is the signed area between the curve and the x-axis from a to b, with area below the axis counted as negative.

The properties of definite integrals are what make EAPCET questions quick. The most powerful is ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx: adding the original and the transformed integral often makes the integrand simplify to a constant. Others include splitting the interval, ∫₋ₐᵃ f(x) dx being twice ∫₀ᵃ for even f and zero for odd f, and properties of periodic functions.

Definite integrals also measure areas. The area between two curves from a to b is ∫ |f(x) − g(x)| dx, found by locating the intersection points and integrating the upper curve minus the lower one. A sum of the form (1/n) Σ f(r/n) becomes ∫₀¹ f(x) dx as n → ∞. Reduction formulas handle powers of sin x and cos x over [0, π/2].

Worked example

EAPCET-style question

The value of ∫25xx+7−x dx\displaystyle\int_{2}^{5} \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{7 - x}}\,dx is

  1. (1)32\dfrac{3}{2}
  2. (2)3
  3. (3)52\dfrac{5}{2}
  4. (4)72\dfrac{7}{2}

Solution

  1. Call the integral II. Here a+b=7a + b = 7, so replace xx by 7−x7 - x: I=∫257−x7−x+x dxI = \displaystyle\int_{2}^{5} \dfrac{\sqrt{7 - x}}{\sqrt{7 - x} + \sqrt{x}}\,dx.
  2. Add the two forms: 2I=∫25x+7−xx+7−x dx=∫251 dx2I = \displaystyle\int_{2}^{5} \dfrac{\sqrt{x} + \sqrt{7 - x}}{\sqrt{x} + \sqrt{7 - x}}\,dx = \int_{2}^{5} 1\,dx.
  3. So 2I=5−2=32I = 5 - 2 = 3.
  4. I=32I = \dfrac{3}{2}.

Answer: 32\dfrac{3}{2}

Shortcuts and results to remember

  • If ff is continuous and ∫abf(x) dx=0\displaystyle\int_{a}^{b} f(x)\,dx = 0, then f(x)=0f(x) = 0 has at least one root in (a,b)(a, b).
  • ∫0π/2sin⁡nxsin⁡nx+cos⁡nx dx=π4\displaystyle\int_{0}^{\pi/2} \dfrac{\sin^{n}x}{\sin^{n}x + \cos^{n}x}\,dx = \dfrac{\pi}{4} for any nn: add the integral to its a+b−xa + b - x form to get ∫1 dx\displaystyle\int 1\,dx.
  • For max⁡{… }\max\{\dots\} or min⁡{… }\min\{\dots\} integrands, find where the functions cross and split the interval there; for [⋅][\cdot] integrands, split where the inside crosses an integer.
  • Area is always positive: when a curve crosses the xx-axis, split at the crossing and add absolute values.
  • When the region is easier to describe with xx as a function of yy (horizontal strips), integrate with respect to yy.

More practice questions

Practice question 1

If f(x)=max⁡{sin⁡x,32,cos⁡x}f(x) = \max\left\{\sin x, \tfrac{\sqrt{3}}{2}, \cos x\right\}, then ∫0π/2f(x) dx\displaystyle\int_{0}^{\pi/2} f(x)\,dx equals

  1. (1)1+3π121 + \dfrac{\sqrt{3}\pi}{12}
  2. (2)1+3π61 + \dfrac{\sqrt{3}\pi}{6}
  3. (3)3π12\dfrac{\sqrt{3}\pi}{12}
  4. (4)2−3π122 - \dfrac{\sqrt{3}\pi}{12}

Solution

  1. On [0,π6]\left[0, \tfrac{\pi}{6}\right], cos⁡x≥32\cos x \ge \tfrac{\sqrt{3}}{2}, so f=cos⁡xf = \cos x
  2. On [π6,π3]\left[\tfrac{\pi}{6}, \tfrac{\pi}{3}\right], both sin⁡x\sin x and cos⁡x\cos x are ≤32\le \tfrac{\sqrt{3}}{2}, so f=32f = \tfrac{\sqrt{3}}{2}
  3. On [π3,π2]\left[\tfrac{\pi}{3}, \tfrac{\pi}{2}\right], sin⁡x≥32\sin x \ge \tfrac{\sqrt{3}}{2}, so f=sin⁡xf = \sin x
  4. Integral =[sin⁡x]0π/6+(32)(π6)+[−cos⁡x]π/3π/2=12+3π12+12= [\sin x]_{0}^{\pi/6} + \left(\tfrac{\sqrt{3}}{2}\right)\left(\tfrac{\pi}{6}\right) + [-\cos x]_{\pi/3}^{\pi/2} = \tfrac{1}{2} + \tfrac{\sqrt{3}\pi}{12} + \tfrac{1}{2}
  5. =1+3π12= 1 + \dfrac{\sqrt{3}\pi}{12}

Answer: 1+3π121 + \dfrac{\sqrt{3}\pi}{12}

Practice question 2

The area of the region {(x,y):y2≤2x and y≥4x−1}\{(x, y) : y^{2} \le 2x \text{ and } y \ge 4x - 1\} is

  1. (1)932\dfrac{9}{32}
  2. (2)1532\dfrac{15}{32}
  3. (3)916\dfrac{9}{16}
  4. (4)524\dfrac{5}{24}

Solution

  1. Solve y2=2xy^{2} = 2x with x=y+14x = \tfrac{y + 1}{4}: y2=y+12y^{2} = \tfrac{y + 1}{2}, so 2y2−y−1=02y^{2} - y - 1 = 0 and y=1y = 1 or y=−12y = -\tfrac{1}{2}
  2. For yy between −12-\tfrac{1}{2} and 1 the line x=y+14x = \tfrac{y + 1}{4} lies to the right of the parabola x=y22x = \tfrac{y^{2}}{2}
  3. Area =∫−1/21[y+14−y22]dy=[y28+y4−y36]−1/21= \displaystyle\int_{-1/2}^{1} \left[\dfrac{y + 1}{4} - \dfrac{y^{2}}{2}\right]dy = \left[\dfrac{y^{2}}{8} + \dfrac{y}{4} - \dfrac{y^{3}}{6}\right]_{-1/2}^{1}
  4. =524−(−796)=2096+796=2796= \tfrac{5}{24} - \left(-\tfrac{7}{96}\right) = \tfrac{20}{96} + \tfrac{7}{96} = \tfrac{27}{96}
  5. =932= \tfrac{9}{32}

Answer: 932\dfrac{9}{32}

Common mistakes to avoid

  • Using the odd-function property on [−a, a] for an integrand that is not odd; check f(−x) = −f(x) before writing zero.
  • Changing the variable by substitution but keeping the old limits; convert the limits to the new variable.
  • Finding the area of a region below the x-axis as a negative number; take the modulus of each part and add.
  • Forgetting that ∫ₐᵇ f(x) dx = −∫ᵦᵃ f(x) dx when the limits are swapped during a calculation.

How to prepare it

Whenever the integrand looks symmetric, try replacing x with a + b − x and adding the two forms.

Once the formulas feel familiar, test them the way the exam does: practise Definite Integrals questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Definite Integrals chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Definite Integrals part of the EAPCET Mathematics syllabus?
Yes. Definite Integrals is taught in Intermediate second year · Paper IIB Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Definite Integrals for EAPCET?
Whenever the integrand looks symmetric, try replacing x with a + b − x and adding the two forms.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Definite Integrals?
Using the odd-function property on [−a, a] for an integrand that is not odd; check f(−x) = −f(x) before writing zero. Changing the variable by substitution but keeping the old limits; convert the limits to the new variable. Finding the area of a region below the x-axis as a negative number; take the modulus of each part and add. Forgetting that ∫ₐᵇ f(x) dx = −∫ᵦᵃ f(x) dx when the limits are swapped during a calculation.