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Q10:
Choose the correct answer in Exercises 9 and 10. 10. If $f(x) = \int_{0}^{x} t \sin t dt$, then $f'(x)$ is

Solution :

Step 1: Application of the Leibniz Integral Rule

To find the derivative of a function defined as a definite integral with variable limits, we employ the Leibniz Integral Rule [Fundamental Theorem of Calculus, Part 1]. For a function defined as:

$f(x) = \int_{g(x)}^{h(x)} \phi(t) dt$

The derivative $f'(x)$ is given by:

$f'(x) = \phi(h(x)) \cdot h'(x) - \phi(g(x)) \cdot g'(x)$

Step 2: Identification of Components

Given the function:

$f(x) = \int_{0}^{x} t \sin t \, dt$

  • $\phi(t) = t \sin t$
  • $h(x) = x$, therefore $h'(x) = \frac{d}{dx}(x) = 1$
  • $g(x) = 0$, therefore $g'(x) = \frac{d}{dx}(0) = 0$

Step 3: Execution of the Differentiation

Applying the formula derived in Step 1:

$f'(x) = \phi(x) \cdot (1) - \phi(0) \cdot (0)$

$f'(x) = (x \sin x) \cdot 1 - 0$

$f'(x) = x \sin x$

Visual Representation of the Integrand

t y=t sin t 0 x

Step 4: Verification via Integration (Alternative Method)

If we evaluate the integral first using Integration by Parts ($\int u dv = uv - \int v du$):

Let $u = t$, $dv = \sin t \, dt \implies du = dt$, $v = -\cos t$

$\int t \sin t \, dt = -t \cos t - \int (-\cos t) \, dt = -t \cos t + \sin t$

Evaluating from $0$ to $x$:

$f(x) = [-t \cos t + \sin t]_{0}^{x} = (-x \cos x + \sin x) - (-0 \cos 0 + \sin 0) = \sin x - x \cos x$

Now, differentiate $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(\sin x - x \cos x)$

$f'(x) = \cos x - [x(-\sin x) + \cos x(1)]$

$f'(x) = \cos x + x \sin x - \cos x$

$f'(x) = x \sin x$

Final Solution: $f'(x) = x \sin x$


More Questions from Class 12 Mathematics Integrals EXERCISE 7.9


CBSE Solutions for Class 12 Mathematics Integrals


Chapters in CBSE - Class 12 Mathematics


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