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3.5 · Q1

Q.Evaluate the following definite integrals:

(i) ∫03f(x) dx\int_0^3 f(x)\,dx where f(x)={x+1x<12xx≥1f(x)=\begin{cases} x+1 & x<1 \\ 2x & x\ge 1 \end{cases}
(ii) ∫01x1−x dx\int_0^1 x\sqrt{1-x}\,dx
(iii) ∫04∣x−2∣ dx\int_0^4 |x-2|\,dx
(iv) ∫15xx+6−x dx\int_1^5 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{6-x}}\,dx
(v) ∫0ax2020x2020+(a−x)2020 dx\int_0^a \frac{x^{2020}}{x^{2020}+(a-x)^{2020}}\,dx
(vi) ∫04(∣x∣+∣x−2∣+∣x−4∣) dx\int_0^4 (|x|+|x-2|+|x-4|)\,dx
(vii) ∫−2211+ex dx\int_{-2}^{2} \frac{1}{1+\sqrt{e^x}}\,dx
(viii) ∫−11x3+∣x∣+1x2+2∣x∣+1 dx\int_{-1}^{1} \frac{x^3+|x|+1}{x^2+2|x|+1}\,dx
(ix) ∫01x(1−x)n dx\int_0^1 x(1-x)^n\,dx
(x) ∫−22(x3+12)4+x2 dx\int_{-2}^{2}\left(x^3+\frac{1}{2}\right)\sqrt{4+x^2}\,dx
(xi) ∫−11log⁡1−x1+x dx\int_{-1}^{1} \log\frac{1-x}{1+x}\,dx
(xii) ∫−1/41/4x21+ex dx\int_{-1/4}^{1/4} \frac{x^2}{1+e^x}\,dx
Puducherry CbseNCERTSubjective· 5mImportance★★★★★
49% · 29/59 Questions
✓ Free question

Twelve definite integrals evaluated using piecewise splitting, substitution, the modulus, the odd/even test and the King property ∫abf(x) dx=∫abf(a+b−x) dx\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx.

King property: ∫abf(x) dx=∫abf(a+b−x) dx\displaystyle\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx.

Symmetry: ∫−aaf(x) dx=0\displaystyle\int_{-a}^{a} f(x)\,dx=0 if ff is odd, =2∫0af(x) dx=2\int_0^a f(x)\,dx if ff is even.

Power rule: ∫xn dx=xn+1n+1\displaystyle\int x^n\,dx=\dfrac{x^{n+1}}{n+1}; ∫x2+a2 dx=x2x2+a2+a22log⁡ ⁣(x+x2+a2)\displaystyle\int\sqrt{x^2+a^2}\,dx=\dfrac{x}{2}\sqrt{x^2+a^2}+\dfrac{a^2}{2}\log\!\left(x+\sqrt{x^2+a^2}\right).

(i) Split at x=1x=1: ∫01(x+1) dx+∫132x dx=[x22+x]01+[x2]13=32+(9−1)=192\displaystyle\int_0^1(x+1)\,dx+\int_1^3 2x\,dx=\Big[\tfrac{x^2}{2}+x\Big]_0^1+\big[x^2\big]_1^3=\tfrac32+(9-1)=\dfrac{19}{2}.

(ii) Put u=1−xu=1-x: ∫01(1−u)u1/2 du=[23u3/2−25u5/2]01=23−25=415\displaystyle\int_0^1(1-u)u^{1/2}\,du=\Big[\tfrac23u^{3/2}-\tfrac25u^{5/2}\Big]_0^1=\tfrac23-\tfrac25=\dfrac{4}{15}.

(iii) ∫02(2−x) dx+∫24(x−2) dx=2+2=4\displaystyle\int_0^2(2-x)\,dx+\int_2^4(x-2)\,dx=2+2=4.

(iv) By the King property (a+b=6a+b=6), if I=∫15xx+6−xdxI=\int_1^5\frac{\sqrt{x}}{\sqrt{x}+\sqrt{6-x}}dx then I=∫156−x6−x+xdxI=\int_1^5\frac{\sqrt{6-x}}{\sqrt{6-x}+\sqrt{x}}dx; adding, 2I=∫151 dx=42I=\int_1^5 1\,dx=4, so I=2I=2.

(v) Same property with a+b=aa+b=a: 2I=∫0a1 dx=a2I=\int_0^a 1\,dx=a, so I=a2I=\dfrac{a}{2}.

(vi) On [0,4][0,4]: ∫04x dx=8\int_0^4 x\,dx=8; ∫04∣x−2∣ dx=4\int_0^4|x-2|\,dx=4; ∫04(4−x) dx=8\int_0^4(4-x)\,dx=8. Sum =8+4+8=20=8+4+8=20.

(vii) With ex=ex/2\sqrt{e^x}=e^{x/2}, f(x)+f(−x)=11+ex/2+ex/21+ex/2=1f(x)+f(-x)=\frac{1}{1+e^{x/2}}+\frac{e^{x/2}}{1+e^{x/2}}=1, so 2I=∫−221 dx=42I=\int_{-2}^{2}1\,dx=4, I=2I=2.

(viii) Denominator =(∣x∣+1)2=(|x|+1)^2 (even). The term x3(∣x∣+1)2\frac{x^3}{(|x|+1)^2} is odd ⇒0\Rightarrow 0; ∣x∣+1(∣x∣+1)2=1∣x∣+1\frac{|x|+1}{(|x|+1)^2}=\frac{1}{|x|+1} is even, so I=2∫01dxx+1=2log⁡2I=2\int_0^1\frac{dx}{x+1}=2\log 2.

(ix) Put u=1−xu=1-x: ∫01(1−u)un du=1n+1−1n+2=1(n+1)(n+2)\int_0^1(1-u)u^n\,du=\frac{1}{n+1}-\frac{1}{n+2}=\dfrac{1}{(n+1)(n+2)}.

(x) x34+x2x^3\sqrt{4+x^2} is odd ⇒0\Rightarrow0; 124+x2\frac12\sqrt{4+x^2} is even, so I=∫02x2+4 dx=[x2x2+4+2log⁡(x+x2+4)]02=22+2log⁡(2+22)−2ln⁡2=22+2log⁡(1+2)I=\int_0^2\sqrt{x^2+4}\,dx=\Big[\tfrac{x}{2}\sqrt{x^2+4}+2\log(x+\sqrt{x^2+4})\Big]_0^2=2\sqrt2+2\log(2+2\sqrt2)-2\ln2=2\sqrt2+2\log(1+\sqrt2).

(xi) f(−x)=log⁡1+x1−x=−f(x)f(-x)=\log\frac{1+x}{1-x}=-f(x) is odd ⇒I=0\Rightarrow I=0.

(xii) Adding f(x)+f(−x)=x21+ex+x21+e−x=x2f(x)+f(-x)=\frac{x^2}{1+e^x}+\frac{x^2}{1+e^{-x}}=x^2, so 2I=∫−1/41/4x2 dx=[x33]−1/41/4=1962I=\int_{-1/4}^{1/4}x^2\,dx=\Big[\tfrac{x^3}{3}\Big]_{-1/4}^{1/4}=\frac{1}{96}, giving I=1192I=\dfrac{1}{192}.

✓Final answer

(i) 192\dfrac{19}{2} (ii) 415\dfrac{4}{15} (iii) 44 (iv) 22 (v) a2\dfrac{a}{2} (vi) 2020 (vii) 22 (viii) 2ln⁡22\ln2 (ix) 1(n+1)(n+2)\dfrac{1}{(n+1)(n+2)} (x) 22+2log⁡(1+2)≈4.592\sqrt2+2\log(1+\sqrt2)\approx4.59 (xi) 00 (xii) 1192\dfrac{1}{192}

Note

For part (xii), our value 1192\dfrac{1}{192} follows from 2I=∫−1/41/4x2 dx=[x33]−1/41/4=1962I=\displaystyle\int_{-1/4}^{1/4}x^2\,dx=\left[\dfrac{x^3}{3}\right]_{-1/4}^{1/4}=\dfrac{1}{96}, hence I=1192I=\dfrac{1}{192}. The official CBSE book's printed answer key gives 196\dfrac{1}{96} for this part — that figure equals 2I2I, so the final division by 22 appears to have been omitted in the printed key. The value above is the correct result of the integral.

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