Category: Definite Integral

Calculus

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.

Let T > 0 be a fixed real number. Suppose f is a continuous function such that for all x ∈ R f(x+T)= f(x)…

Q: Let T > 0 be a fixed real number. Suppose f is a continuous function such that for all x ∈ R f(x+T)= f(x). If $I = \int_{0}^{T} f(x) dx $ , then the value of $ \int_{0}^{3+3T} f(2x) dx $  is

(A) $\frac{3}{2}I $

(B) 2I

(C) 3I

(D) 6I

Sol: Let $\large L = \int_{0}^{3 + 3T} f(2x) dx $

Put 2x = t so that

$\large L = \frac{1}{2}\int_{0}^{6 + 6T} f(t) dt $

$\large L = \frac{6}{2}\int_{0}^{T} f(t) dt = 3 I$

Hence (C) is the correct answer.

The value of $\int_{0}^{sin^2 x} sin^{-1}(\sqrt{t}) dt + \int_{0}^{cos^2 x} cos^{-1}(\sqrt{t}) dt $

Q: The value of $\int_{0}^{sin^2 x} sin^{-1}(\sqrt{t}) dt + \int_{0}^{cos^2 x} cos^{-1}(\sqrt{t}) dt $ is

(A) π/4

(B) 0

(C) π/2

(D) none of these

Sol: Let $\large f(x) = \int_{0}^{sin^2 x} sin^{-1}(\sqrt{t}) dt + \int_{0}^{cos^2 x} cos^{-1}(\sqrt{t}) dt $

Here f'(x) = 0

⇒  f(x) = c (a constant).

But $\large f(\pi/4) = \int_{0}^{1/2} (sin^{-1}(\sqrt{t} + cos^{-1}(\sqrt{t}) dt $

$\large = \int_{0}^{1/2} \frac{\pi}{2} dt = \frac{\pi}{4} $

Hence (A) is the correct answer.