Definite Integral

Calculus

$\int_{0}^{\pi} x f(sin x) dx $ is equal to

Q: $\displaystyle \int_{0}^{\pi} x f(sin x) dx $ is equal to

(A) $ \frac{\pi}{2} \int_{0}^{\pi} f(sin x) dx $

(B) $ \pi \int_{0}^{\pi/2} f(sin x) dx $

(C) $ 2 \pi \int_{0}^{\pi/2} f(sin x) dx $

(D) none of these

Ans: (A) , (B)

Sol: Let $\displaystyle I = \int_{0}^{\pi} x f(sin x) dx $

$\displaystyle = \int_{0}^{\pi} (\pi – x ) f sin (\pi-x) dx $

$\displaystyle I = \pi \int_{0}^{\pi} f(sin x) dx – I $

$\displaystyle 2 I = \pi \int_{0}^{\pi} f(sin x) dx $

$\displaystyle I = \frac{\pi}{2} \int_{0}^{\pi} f(sin x) dx $

$\displaystyle I = 2 \times \frac{\pi}{2} \int_{0}^{\pi/2} f(sin x) dx $

$\displaystyle I = \pi \int_{0}^{\pi/2} f(sin x) dx $

Hence (A) and (B) are the correct answers.

If $ I = \int_{0}^{\pi/2} \frac{dx}{\sqrt{1 + sin^3 x}} $ , then

Q: If $\displaystyle I = \int_{0}^{\pi/2} \frac{dx}{\sqrt{1 + sin^3 x}} $ , then

(A) 0 < I < 1

(B) $ I > \frac{\pi}{2\sqrt{2}}$

(C) I < √2 π

(D) I > 2 π

Ans: (B) , (C)

Sol: Since x ∈ [0 , π/2 ]

⇒ 1 ≤ 1 + sin3 x  ≤  2

$\frac{1}{\sqrt{2}} \le \frac{1}{\sqrt{1 + sin^3 x}} \le 1 $

$ \int_{0}^{\pi/2} \frac{1}{\sqrt{2}} dx \le \int_{0}^{\pi/2} \frac{1}{\sqrt{1 + sin^3 x}} dx \le \int_{0}^{\pi/2} 1 dx $

$\frac{\pi}{2 \sqrt{2}} \le I \le \frac{\pi}{2}$

Hence (B) and (C) are the correct answers.

The value of $ I = \int_{0}^{3}([x] + [x + \frac{1}{3}] + [x+\frac{2}{3}])dx$

Q: The value of $\large I = \int_{0}^{3}([x] + [x + \frac{1}{3}] + [x+\frac{2}{3}])dx$

where [.] denotes the greatest integer function, is equal to;

(A) 10

(B) 11

(C) 12

(D) none of these

Ans: (C)

Sol: Let $\displaystyle I = \int_{0}^{1/3} 0.dx + \int_{1/3}^{2/3} 1 .dx + \int_{2/3}^{1} 2.dx + \int_{1}^{4/3} 3 .dx + \int_{4/3}^{5/3} 4.dx $

$\displaystyle + \int_{5/3}^{6/3} 5.dx + \int_{6/3}^{7/3} 6 .dx + \int_{7/3}^{8/3} 7 .dx + \int_{8/3}^{9/3} 8 .dx $

$\displaystyle I = \frac{1}{3}(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 ) $

= 12

Hence (C) is the correct answer.

Let g (x) = sin x + cos x and f (x) = ….

Q: Let g (x) = sin x + cos x and $ \large f(x) = \left\{\begin{array}{ll} \frac{|x|}{x} \; , x \ne 0 \\ 0 \; , x = 0 \end{array} \right. $ then value of $\large \int_{-\pi/4}^{2\pi}f(g(x))dx $ is equal to

(A) π/2

(B) π/4

(C) π

(D) none of these

Sol: $ \large f(g(x)) = \left\{\begin{array}{lll} 1 \; ,  sinx + cosx  > 0 \\ -1 \; , sinx + cosx  < 0  \\ 0 \; , sinx + cosx  = 0\end{array} \right. $

Now sin x + cos x > 0

$\large \sqrt{2}sin(x+\frac{\pi}{4}) > 0 $

$\large \int_{-\pi/4}^{2\pi}f(g(x))dx $

$\large = \int_{-\pi/4}^{3\pi/4}1 dx + \int_{3\pi/4}^{7\pi/4}-1 dx + \int_{7\pi/4}^{2\pi}1 dx $

$\large = \pi – \pi + \frac{\pi}{4} = \frac{\pi}{4} $

Hence (B) is the correct answer.

If for non-zero x , $ 3 f(x) + 4 f(\frac{1}{x}) = \frac{1}{x} – 10 $ then $\int_{2}^{3}f(x) dx $ is equal to

Q: If for non-zero x , $\large 3 f(x) + 4 f(\frac{1}{x}) = \frac{1}{x} – 10 $ then $\large \int_{2}^{3}f(x) dx $ is equal to

(A) $\frac{4}{7} ln\frac{2}{3}$

(B) $\frac{3}{7} ln\frac{3}{2}$

(C) $\frac{3}{7} ln\frac{2}{3}$

(D) None of these

Sol: $\large 3 f(x) + 4 f(\frac{1}{x}) = \frac{1}{x} – 10 $

replace x by 1/x

$\large 3 f(\frac{1}{x}) + 4 f(x) = x – 10 $

solving (1) and (2)

9f(x) – 16f (x) = 3/x – 30 – 4x + 40

– 7f (x) = 10 + 3/x – 4x

$\large f(x) = -\frac{10}{7}-\frac{3}{7x} + \frac{4x}{7} $

$\large \int_{2}^{3}f(x) dx = \int_{2}^{3} (-\frac{10}{7}-\frac{3}{7x} + \frac{4x}{7}) dx $

$\large [-\frac{10}{7}x -\frac{3}{7}ln|x| + \frac{4x^2}{14}]_{2}^{3}$

$\large = \frac{3}{7}ln\frac{2}{3}$

Hence (C) is the correct answer.