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Worked Examples · Example 11

Q.Evaluate the following definite integrals:

(a) ∫03x3 dx\int_0^3 x^3\,dx
(b) ∫0111+x2 dx\int_0^1 \frac{1}{\sqrt{1+x^2}}\,dx
(c) ∫14x(x+1)(x+4) dx\int_1^4 \frac{x}{(x+1)(x+4)}\,dx
(d) ∫02x2+4 dx\int_0^2 \sqrt{x^2+4}\,dx
(e) ∫01xex dx\int_0^1 xe^x\,dx
(f) ∫35x2(x−1)(x−2) dx\int_3^5 \frac{x^2}{(x-1)(x-2)}\,dx
Chandigarh CbseNCERTSubjective· 5mImportance★★★★★
19% · 11/59 Questions
✓ Free question

Find each antiderivative (power rule, standard radical form, partial fractions, by parts), then evaluate between the limits.

∫abf(x) dx=F(b)−F(a),∫dxx2+a2=log⁡ ⁣∣x+x2+a2∣,∫x2+a2 dx=x2x2+a2+a22log⁡ ⁣∣x+x2+a2∣.\int_a^b f(x)\,dx=F(b)-F(a),\quad \int\frac{dx}{\sqrt{x^2+a^2}}=\log\!\big|x+\sqrt{x^2+a^2}\big|,\quad \int\sqrt{x^2+a^2}\,dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log\!\big|x+\sqrt{x^2+a^2}\big|.

Steps

  1. (a) ∫03x3dx=[x44]03=814=20.25.\displaystyle\int_0^3 x^3dx=\Big[\frac{x^4}{4}\Big]_0^3=\frac{81}{4}=20.25.

  2. (b) ∫01dx1+x2=[log⁡(x+1+x2)]01=log⁡(1+2)−log⁡1=log⁡(1+2)≈0.881.\displaystyle\int_0^1\frac{dx}{\sqrt{1+x^2}}=\Big[\log\big(x+\sqrt{1+x^2}\big)\Big]_0^1=\log(1+\sqrt2)-\log 1=\log(1+\sqrt2)\approx0.881.

  3. (c) Partial fractions: x(x+1)(x+4)=−1/3x+1+4/3x+4\dfrac{x}{(x+1)(x+4)}=\dfrac{-1/3}{x+1}+\dfrac{4/3}{x+4} (at x=−1x=-1, A=−13A=-\tfrac13; at x=−4x=-4, B=43B=\tfrac43).

∫14=[−13log⁡∣x+1∣+43log⁡∣x+4∣]14=(−13ln⁡5+43ln⁡8)−(−13ln⁡2+43ln⁡5).\int_1^4=\Big[-\tfrac13\log|x+1|+\tfrac43\log|x+4|\Big]_1^4=\Big(-\tfrac13\ln5+\tfrac43\ln8\Big)-\Big(-\tfrac13\ln2+\tfrac43\ln5\Big).

Using ln⁡8=3ln⁡2\ln8=3\ln2: =4ln⁡2+13ln⁡2−53ln⁡5=13ln⁡2−5ln⁡53≈0.321.=4\ln2+\tfrac13\ln2-\tfrac53\ln5=\dfrac{13\ln2-5\ln5}{3}\approx0.321.

  1. (d) With a=2a=2: ∫02x2+4 dx=[x2x2+4+2log⁡(x+x2+4)]02.\displaystyle\int_0^2\sqrt{x^2+4}\,dx=\Big[\frac{x}{2}\sqrt{x^2+4}+2\log\big(x+\sqrt{x^2+4}\big)\Big]_0^2. At x=2x=2: 228+2log⁡(2+8)=22+2log⁡(2+22)\tfrac22\sqrt8+2\log(2+\sqrt8)=2\sqrt2+2\log(2+2\sqrt2). At x=0x=0: 0+2ln⁡20+2\ln2.

=22+2log⁡2+222=22+2log⁡(1+2)≈2.828+1.763=4.591.=2\sqrt2+2\log\frac{2+2\sqrt2}{2}=2\sqrt2+2\log(1+\sqrt2)\approx2.828+1.763=4.591.

  1. (e) By parts, ∫xexdx=ex(x−1)\int xe^{x}dx=e^{x}(x-1):

∫01xexdx=[ex(x−1)]01=e(0)−1⋅(−1)=0+1=1.\int_0^1 xe^{x}dx=\big[e^{x}(x-1)\big]_0^1=e(0)-1\cdot(-1)=0+1=1.

  1. (f) Improper: x2(x−1)(x−2)=1+3x−2(x−1)(x−2)=1−1x−1+4x−2\dfrac{x^2}{(x-1)(x-2)}=1+\dfrac{3x-2}{(x-1)(x-2)}=1-\dfrac{1}{x-1}+\dfrac{4}{x-2} (at x=1x=1, A=−1A=-1; at x=2x=2, B=4B=4).

∫35=[x−log⁡∣x−1∣+4log⁡∣x−2∣]35=(5−ln⁡4+4ln⁡3)−(3−ln⁡2+0).\int_3^5=\Big[x-\log|x-1|+4\log|x-2|\Big]_3^5=\big(5-\ln4+4\ln3\big)-\big(3-\ln2+0\big).

Using ln⁡4=2ln⁡2\ln4=2\ln2: =2−2ln⁡2+ln⁡2+4ln⁡3=2−ln⁡2+4ln⁡3≈5.701.=2-2\ln2+\ln2+4\ln3=2-\ln2+4\ln3\approx5.701.

✓Final answer

  1. 20.2520.25;
  2. log⁡(1+2)≈0.881\log(1+\sqrt2)\approx0.881;
  3. 13ln⁡2−5ln⁡53≈0.321\dfrac{13\ln2-5\ln5}{3}\approx0.321;
  4. 22+2log⁡(1+2)≈4.5912\sqrt2+2\log(1+\sqrt2)\approx4.591; (e) 11; (f) 2−ln⁡2+4ln⁡3≈5.7012-\ln2+4\ln3\approx5.701.

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