Newton Leibniz Theorem provides the formula for the differentiation of a definite integral whose limits are functions of the differential variable. This is also known as differentiation under the integral sign. Differentiation and integration are important topics for the JEE Main exam. Using Newton Leibniz Theorem, students can easily solve questions on the differentiation of a definite integral when limits are functions of the differential variable.

Let F(x) be a function such that f(x) is its derivative. Then, ∫_a^bf(x) dx = F(b) – F(a).

When the limits of a definite integral are functions of t, and the integrands are functions of x or vice versa, we can use Newton Leibniz Theorem to find the derivative of the definite integral.

Newton Leibnitz Formula for Limits

\begin{array}{l} \frac{d}{d x} (\int_{h (x)}^{g (x)} f (t) d t) = f (g (x)) \times g^{'} (x) - f (h (x)) \times h^{'} (x) \end{array}

= f(upper limit) × derivative of upper limit – f(lower limit)× derivative of lower limit

Solved Examples

Example 1:

Differentiate ∫₀^x cos t dt.

Solution:

cos t is not a function of x.

We use the formula

\begin{array}{l} \frac{d}{d x} (\int_{h (x)}^{g (x)} f (t) d t) = f (g (x)) \times g^{'} (x) - f (h (x)) \times h^{'} (x) \end{array}

Here, h(x) = x and g(x) = 0

f(t) = cos t

So, d/dx ∫₀^x cos t dt = cos x × (d/dx)(x) – cos 0 (d/dx) 0

= cos x – 0

= cos x

Example 2:

\begin{array}{l} Evaluate \frac{d}{d x} [\int_{\frac{1}{x}}^{\sqrt{x}} \cos t^{2} d t] \end{array}

Solution:

Given

\begin{array}{l} \frac{d}{d x} [\int_{\frac{1}{x}}^{\sqrt{x}} \cos t^{2} d t] \end{array}

= cos (√x)² (d/dx)√x – cos(1/x)²(d/dx)(1/x)

= cos x. (1/2√x) + (1/x²) cos (1/x²)

= (cos x)/2√x + (1/x²) cos (1/x²)

Example 3:

If ∫₀^xf(t) dt = x + ∫_x¹t f(t) dt, then the value of f(1) is

(a) 1/2

(b) 1

(c) 0

(d) none of these

Solution:

Given ∫₀^xf(t) dt = x + ∫_x¹t f(t) dt

Differentiate both sides w.r.t. x

(d/dx) ∫₀^xf(t) dt = (d/dx) [x + ∫_x¹t f(t) dt]

Using Newton Leibniz Theorem, we get,

f(x)×1 – f(0)×0 = 1 + 1×f(1)×0 – x f(x)×1

f(x) = 1+ 0 – x f(x)

f(x) + x f(x) = 1

f(x)(1+x) = 1

f(x) = 1/(1+x)

So f(1) = 1/(1+1)

= 1/2

Hence, option (a) is the answer.

Example 4:

\begin{array}{l} \int_{0}^{t^{2}} x f (x) = \frac{2}{5} t^{5}, t > 0, then f (\frac{4}{25}) = \end{array}

(a) 2/5

(b) 5/2

(c) 0

(d) none of these

Solution:

\begin{array}{l} \int_{0}^{t^{2}} x f (x) = \frac{2}{5} t^{5} \end{array}

Here, the integrand is a function of x, limit is a function of t. We can use Newton Leibnitz formula for limits to solve this.

Differentiate both sides with respect to t.

\begin{array}{l} \frac{d}{d t} (\int_{0}^{t^{2}} x f (x)) = \frac{d}{d t} \frac{2}{5} t^{5} \end{array}

t² ×f(t²) ×2t – 0 = 2t⁴

f(t²) = t

So f(4/25) = 2/5

Hence, option (a) is the answer.

Question 5:

If f(x) be a function defined by f(x) = ∫₁^x t(t² – 3t + 2)dt, 1 ≤ x ≤ 3, find the range of f(x).

Solution:

Given f(x) = ∫₁^x t(t² – 3t + 2)dt

We can see that the limits are functions of x, and the integrand is a function of t.

We can use Newton Leibnitz’s formula for limits to solve this.

Differentiate both sides with respect to x.

f’(x) = (d/dx) ∫₁^x t(t² – 3t + 2)dt

= f(upper limit) × derivative of upper limit – f(lower limit)× derivative of the lower limit

= x (x² – 3x + 2) – 1(1 – 3 + 2)0

= x (x² – 3x + 2)

= x (x – 2) (x – 1)

f(x) is decreasing in [1,2] and increasing in [2,3]

f(2) = ∫₁² t(t² – 3t + 2)dt

= ∫₁² (t³ – 3t² + 2t)dt

= [t⁴/4 – t³ + t²]₁²

= [4 – 8 + 4 ] – [¼ – 1 + 1]

= -1/4

f(1) = 0

f(3) = (81/4 – 27 + 9) – (¼ – 1 + 1)

= 80/4 – 18

= 20 – 18

= 2

The range of f(x) is [-¼, 2].

Newton Leibnitz’s Theorem – Concept Video

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