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The maximum value of int (-pi//2) ^(2pi/...

The maximum value of `int _(-pi//2) ^(2pi//2) sin x. f (x) dx,` subject to the condition `|f (x) le 5` is M, then `M/10is equal to :

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To solve the problem of finding the maximum value of the integral \[ M = \int_{-\frac{\pi}{2}}^{\frac{2\pi}{2}} \sin x \cdot f(x) \, dx \] subject to the condition \(|f(x)| \leq 5\), we can follow these steps: ### Step 1: Understand the Integral and the Condition The integral is defined from \(-\frac{\pi}{2}\) to \(\pi\) (since \(\frac{2\pi}{2} = \pi\)). The function \(f(x)\) is constrained such that its absolute value does not exceed 5. ### Step 2: Substitute the Maximum Value of \(f(x)\) To maximize the integral, we can take \(f(x) = 5\) for the entire interval where \(\sin x\) is positive. The sine function is positive between \(0\) and \(\pi\). ### Step 3: Break Down the Integral The integral can be split into two parts: \[ M = \int_{-\frac{\pi}{2}}^{0} \sin x \cdot f(x) \, dx + \int_{0}^{\pi} \sin x \cdot f(x) \, dx \] For the first part, since \(\sin x\) is negative, we can take \(f(x) = -5\) to maximize the negative contribution: \[ M = \int_{-\frac{\pi}{2}}^{0} \sin x \cdot (-5) \, dx + \int_{0}^{\pi} \sin x \cdot 5 \, dx \] ### Step 4: Evaluate the Integrals Now we evaluate each integral separately. 1. **First Integral:** \[ \int_{-\frac{\pi}{2}}^{0} \sin x \cdot (-5) \, dx = -5 \int_{-\frac{\pi}{2}}^{0} \sin x \, dx \] The integral of \(\sin x\) is \(-\cos x\): \[ = -5 \left[-\cos x \right]_{-\frac{\pi}{2}}^{0} = -5 \left[-\cos(0) + \cos\left(-\frac{\pi}{2}\right)\right] = -5 \left[-1 + 0\right] = 5 \] 2. **Second Integral:** \[ \int_{0}^{\pi} \sin x \cdot 5 \, dx = 5 \int_{0}^{\pi} \sin x \, dx \] Again, using the integral of \(\sin x\): \[ = 5 \left[-\cos x \right]_{0}^{\pi} = 5 \left[-\cos(\pi) + \cos(0)\right] = 5 \left[1 + 1\right] = 10 \] ### Step 5: Combine the Results Now we can combine the results of both integrals: \[ M = 5 + 10 = 15 \] ### Step 6: Find \(M/10\) Finally, we need to find \(M/10\): \[ \frac{M}{10} = \frac{15}{10} = 1.5 \] ### Final Answer Thus, the value of \(M/10\) is: \[ \frac{M}{10} = 1.5 \]
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VIKAS GUPTA (BLACK BOOK) ENGLISH-INDEFINITE AND DEFINITE INTEGRATION-EXERCISE (SUBJECTIVE TYPE PROBLEMS)
  1. Let int ( 1) ^(sqrt5)(x ^(2x ^(2)+1) +ln "("x ^(2x ^(2x ^(2)+1))")")dx...

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  2. If int (dx )/(cos ^(3) x-sin ^(3))=A tan ^(-1) (f (x)) +bln |(sqrt2+f ...

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  3. Find the value of |a| for which the area of triangle included between ...

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  4. Let I = int (0) ^(pi) x ^(6) (pi-x) ^(8)dx, then (pi ^(15))/((""^(15) ...

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  5. If I = int (0) ^(100) (sqrtx)dx, then the value of (9I)/(155) is:

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  6. Let I(n) = int (0)^(pi) (sin (n + (1)/(2))x )/(sin ((x)/(2)))dx where ...

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  7. If M be the maximum value of 72 int (0) ^(y) sqrt(x ^(4) +(y-y^(2))^(2...

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  8. Find the number points where f (theta) = int (-1)^(1) (sin theta dx )/...

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  9. underset(nrarroo)lim[(1)/(sqrtn)+(1)/(sqrt(2n))+(1)/(sqrt(3n))+...+(1)...

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  10. The maximum value of int (-pi//2) ^(2pi//2) sin x. f (x) dx, subject t...

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  11. Given a function g, continous everywhere such that g (1)=5 and int (0)...

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  12. If f (n)= 1/pi int (0) ^(pi//2) (sin ^(2) (n theta) d theta)/(sin ^(2)...

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  13. Let f (2-x) =f (2+xand f (4-x )= f (4+x). Function f (x) satisfies int...

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  14. Let l (n) =int (-1) ^(1) |x|(1+ x+ (x ^(2))/(2 ) +(x ^(3))/(3) + ........

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  15. Let lim ( x to oo) n ^((1)/(2 )(1+(1 )/(n))). (1 ^(1) . 2 ^(2) . 3 ^(3...

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  16. If int (a )^(b) |sin x |dx =8 and int (0)^(a+b) |cos x| dx=9 then the ...

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  17. If f(x),g(x),h(x) and phi(x) are polynomial in x, (int1^x f(x) h(x) dx...

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  18. If int (0)^(2)(3x ^(2) -3x +1) cos (x ^(3) -3x ^(2)+4x -2) dx = a sin ...

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  19. let f (x) = int (0) ^(x) e ^(x-y) f'(y) dy - (x ^(2) -x+1)e ^(x) Fin...

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  20. For a positive integer n, let I (n) =int (-pi)^(pi) ((pi)/(2) -|x|) co...

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