For a, b R and a < b, _a^b (f(x))/(f(x) + f(a + b - x)) dx = Option 1: a - b Option 2: (a + b)/2 Option 3: 2b Option 4: (b - a)/2

Answer: (b - a)/2. Solution: Let I = _a^b (f(x))/(f(x) + f(a+b-x)) dx. Using property _a^b g(x) dx = _a^b g(a+b-x) dx: I = _a^b (f(a+b-x))/(f(a+b-x) + f(x)) dx. Adding: 2I = _a^b 1 dx = b - a. So I = (b-a)/2.

Calculus

Integrals

Medium

For $a, b \in R$ and $a < b$ , $\int_{a}^{b} \frac{f(x)}{f(x) + f(a + b - x)} \, dx =$

$a - b$

$\frac{a + b}{2}$

$2b$

$\frac{b - a}{2}$

✅ Correct Option: 4

Let $I = \int_{a}^{b} \frac{f(x)}{f(x) + f(a+b-x)} \, dx$ .

Using property $\int_{a}^{b} g(x) \, dx = \int_{a}^{b} g(a+b-x) \, dx$ :

$I = \int_{a}^{b} \frac{f(a+b-x)}{f(a+b-x) + f(x)} \, dx$ .

Adding: $2I = \int_{a}^{b} 1 \, dx = b - a$ .

So $I = \frac{b-a}{2}$ .

2025: 3 June Shift 2

$\int e^{2x}(\sin x + \frac{1}{2}\cos x) dx$ is equal to

2026: 29 May Shift 2

$\int_1^2 \frac{dx}{(x+1)(x+2)} =$

2025: 15 May Shift 2

Match List-I with List-II

[.] denotes the greatest integer function.

List-I	List-II
(A) $\int_0^3 [x]dx$	(I) $\frac{1}{2}$
(B) $\int_0^1 [2x]dx$	(II) 1
(C) $\int_0^1 [3x]dx$	(III) $\frac{3}{2}$
(D) $\int_0^1 [4x]dx$	(IV) 3

Choose the correct answer from the options given below:

2025: 13 May Shift 2

$\int_{-\frac{5}{2}}^{\frac{5}{2}} |x| dx$ is equal to

2025: 30 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $\int_0^1 \frac{x^2}{1 + x^3} dx$	(I) 0
(B) $\int_0^\pi 3\sin x dx$	(II) $2\log_e\left(\frac{3}{2}\right)$
(C) $\int_{-1}^1 \sin^5 x \cos^6 x dx$	(III) 6
(D) $\int_2^3 \frac{4}{x^2 - 1} dx$	(IV) $\frac{1}{3}\log_e 2$

Choose the correct answer from the options given below:

2026: 22 May Shift 1

$\int_{4}^{10} \frac{\log(x^2)}{\log(x^2)+\log(196-28x+x^2)}dx$ is equal to : (Consider $\log_e x = \log x$ )

For a,b∈Ra, b \in Ra,b∈R and a<ba < ba<b, ∫abf(x)f(x)+f(a+b−x) dx=\int_{a}^{b} \frac{f(x)}{f(x) + f(a + b - x)} \, dx =∫ab​f(x)+f(a+b−x)f(x)​dx=

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

For $a, b \in R$ and $a < b$ , $\int_{a}^{b} \frac{f(x)}{f(x) + f(a + b - x)} \, dx =$