JEE Main 2013

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: The intercepts on $x$ -axis made by tangents to the curve, $y = \int_{0}^{x} ∣ t ∣ d t, x \in R$ , which are parallel to the line $y = 2 x$ , are equal to

Q: Why do we get two different tangent lines? Is the curve not just one function?

The curve $y = \int_{0}^{x} |t| dt$ is indeed a single continuous function, but it is defined piecewise. For $x \ge 0$, $y = \frac{x^2}{2}$, and for $x < 0$, $y = -\frac{x^2}{2}$. Because the slope $|x|$ equals $2$ at both $x=2$ and $x=-2$, there are two distinct points on the curve where the tangent line has a slope of $2$. Hence, two distinct tangent lines.

Q: I am confused about the integral for $x < 0$. How did we get $y = -2$ for $x = -2$?

Excellent catch. When $x = -2$, the integral is $\int_{0}^{-2} |t| dt$. Since the interval is from $0$ to $-2$, $t$ is negative, so $|t| = -t$. The integral becomes $\int_{0}^{-2} (-t) dt$. Integrating $-t$ gives $-\frac{t^2}{2}$. Evaluating this from $0$ to $-2$ yields $-\frac{(-2)^2}{2} - 0 = -2$. The negative sign is crucial here!

Select Answer:

Visualized Solution

Visualizing the Curve $y = \int_{0}^{x} ∣ t ∣ d t$

The given curve is defined by an integral: $y = \int_{0}^{x} ∣ t ∣ d t$ .
We need to find tangents to this curve that are parallel to the line $y = 2 x$ .
Parallel lines have the same slope, so the slope of our tangents must be $2$ .

The Newton-Leibniz Rule

To find the slope of the curve, we need its derivative, $\frac{d y}{d x}$ .
We use the Newton-Leibniz Rule for differentiating under the integral sign:
$\frac{d}{d x} \int_{a (x)}^{b (x)} f (t) d t = f (b (x)) \cdot b^{'} (x) - f (a (x)) \cdot a^{'} (x)$

Differentiating the Integral

Applying the rule to $y = \int_{0}^{x} ∣ t ∣ d t$ :
$\frac{d y}{d x} = ∣ x ∣ \cdot \frac{d}{d x} (x) - ∣0∣ \cdot \frac{d}{d x} (0)$
$\frac{d y}{d x} = ∣ x ∣ \cdot 1 - 0 = ∣ x ∣$

Setting the Slope Condition

The tangents are parallel to $y = 2 x$ , so their slope is $m = 2$ .
We equate the derivative to the required slope:
$∣ x ∣ = 2$
Solving this absolute value equation gives two possible $x$ -coordinates:
$x = 2$ or $x = - 2$

Finding the First Point of Tangency

Let's find the $y$ -coordinate for $x = 2$ .
Substitute $x = 2$ into the original curve equation:
$y = \int_{0}^{2} ∣ t ∣ d t = \int_{0}^{2} t d t$
$y = [\frac{t ^{2}}{2}]_{0}^{2} = \frac{4}{2} = 2$
The first point of tangency is $(2, 2)$ .

Finding the Second Point of Tangency

Now, let's find the $y$ -coordinate for $x = - 2$ .
Substitute $x = - 2$ into the original curve equation:
$y = \int_{0}^{- 2} ∣ t ∣ d t = \int_{0}^{- 2} (- t) d t$
$y = [- \frac{t ^{2}}{2}]_{0}^{- 2} = - (\frac{( - 2 ) ^{2}}{2} - 0) = - 2$
The second point of tangency is $(- 2, - 2)$ .

Equation of the First Tangent

We use the point-slope form: $y - y_{1} = m (x - x_{1})$ .
For the point $(2, 2)$ and slope $m = 2$ :
$y - 2 = 2 (x - 2)$
$y - 2 = 2 x - 4$
$y = 2 x - 2$

Equation of the Second Tangent

For the point $(- 2, - 2)$ and slope $m = 2$ :
$y - (- 2) = 2 (x - (- 2))$
$y + 2 = 2 (x + 2)$
$y + 2 = 2 x + 4$
$y = 2 x + 2$

Calculating the $x$ -intercepts

The question asks for the $x$ -intercepts of these tangents.
To find the $x$ -intercept, we set $y = 0$ .
For $y = 2 x - 2$ : $0 = 2 x - 2 \Rightarrow 2 x = 2 \Rightarrow x = 1$ .
For $y = 2 x + 2$ : $0 = 2 x + 2 \Rightarrow 2 x = - 2 \Rightarrow x = - 1$ .

Final Conclusion

The $x$ -intercepts made by the tangents are $1$ and $- 1$ .
This can be written compactly as $\pm 1$ .
Final Answer: The intercepts are $\pm 1$ .

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The Sigma Insight: Newton-Leibniz & Reduction Formulas

Solution Diagram

Analyzing the Setup

The curve is defined by the integral function:

y = \int_{0}^{x} ∣ t ∣ d t

The function

∣ t ∣

represents a V-shaped graph. Integrating this from

0

x

calculates the signed area under this graph.

For

x \geq 0

, the area is the integral of

t

, which yields:

y = \frac{x ^{2}}{2}

For

x < 0

, the area accumulates as the integral of

- t

, which yields:

y = - \frac{x ^{2}}{2}

This results in a continuous curve composed of two parabolic segments joined at the origin.

The Power of Newton-Leibniz

We seek tangents to this curve that are parallel to the line

y = 2 x

. Since parallel lines share the same slope, our target slope is

m = 2

To find the slope of the curve at any point, we apply the Newton-Leibniz rule. This rule states that the derivative of an integral with respect to its upper limit is the integrand evaluated at that limit:

\frac{d y}{d x} = ∣ x ∣

This confirms that the slope of the curve at any point

x

is exactly the absolute value of

x

Finding the Points of Tangency

We set the derivative equal to our target slope:

∣ x ∣ = 2

This equation yields two points of interest:

x = 2

and

x = - 2

. We now calculate the corresponding

y

-coordinates by evaluating the original integral at these points.

For

x = 2

y = \int_{0}^{2} t d t = [\frac{t ^{2}}{2}]_{0}^{2} = 2

For

x = - 2

y = \int_{0}^{- 2} (- t) d t = [- \frac{t ^{2}}{2}]_{0}^{- 2} = - 2

The points of tangency are

(2, 2)

and

(- 2, - 2)

The Final Intercepts

Using the point-slope form

y - y_{1} = m (x - x_{1})

with

m = 2

, we construct the equations for the tangent lines.

For the point

(2, 2)

y - 2 = 2 (x - 2) \Rightarrow y = 2 x - 2

For the point

(- 2, - 2)

y - (- 2) = 2 (x - (- 2)) \Rightarrow y = 2 x + 2

To find the

x

-intercepts, we set

y = 0

in both equations. For

y = 2 x - 2

, we find

x = 1

. For

y = 2 x + 2

, we find

x = - 1

The final

x

-intercepts are $\pm 1$ .

Common Traps & FAQs

Wait, why is the derivative of

\int_{0}^{x} ∣ t ∣ d t

simply

∣ x ∣

? Shouldn't there be more to it?

That is a great question! The Newton-Leibniz rule (or the Fundamental Theorem of Calculus) states that if

F (x) = \int_{a}^{x} f (t) d t

, then

F^{'} (x) = f (x)

. Here, our integrand is

f (t) = ∣ t ∣

. So, when we differentiate with respect to

x

, we simply replace

t

with

x

. It is that elegant! The absolute value sign stays because it is part of the function

f (t)

Why do we get two different tangent lines? Is the curve not just one function?

The curve

y = \int_{0}^{x} ∣ t ∣ d t

is indeed a single continuous function, but it is defined piecewise. For

x \geq 0

y = \frac{x ^{2}}{2}

, and for

x < 0

y = - \frac{x ^{2}}{2}

. Because the slope

∣ x ∣

equals

2

at both

x = 2

and

x = - 2

, there are two distinct points on the curve where the tangent line has a slope of

2

. Hence, two distinct tangent lines.

I am confused about the integral for

x < 0

. How did we get

y = - 2

for

x = - 2

Excellent catch. When

x = - 2

, the integral is

\int_{0}^{- 2} ∣ t ∣ d t

. Since the interval is from

0

- 2

t

is negative, so

∣ t ∣ = - t

. The integral becomes

\int_{0}^{- 2} (- t) d t

. Integrating

- t

gives

- \frac{t ^{2}}{2}

. Evaluating this from

0

- 2

yields

- \frac{( - 2 ) ^{2}}{2} - 0 = - 2

. The negative sign is crucial here!

Animated Solution for Mathematics - Definite Integration: The intercepts on $x$ -axis made by tangents to the curve, $y = \int_{0}^{x} ∣ t ∣ d t, x \in R$ , which are parallel to the line $y = 2 x$ , are equal to

Select Answer:

Visualized Solution

Visualizing the Curve $y = \int_{0}^{x} ∣ t ∣ d t$

The Newton-Leibniz Rule

Differentiating the Integral

Setting the Slope Condition

Finding the First Point of Tangency

Finding the Second Point of Tangency

Equation of the First Tangent

Equation of the Second Tangent

Calculating the $x$ -intercepts

Final Conclusion

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Power of Newton-Leibniz

Finding the Points of Tangency

The Final Intercepts

Similar Questions

The total number of distinct $x \in [0, 1]$ for which $\int_{0}^{x} \frac{t ^{2}}{1 + t ^{4}} d t = 2 x - 1$ is

Let $f (x) = \int_{1}^{x} 2 - t^{2} d t$ . Then the real roots of the equation $x^{2} - f^{'} (x) = 0$ are

Let $F (x) = \int_{x}^{x^{2} + \frac{π}{6}} 2 cos^{2} t (d t)$ for all $x \in R$ and $f : [0, 1/2] \to [0, \infty)$ be a continuous function. For $α \in [0, 1/2]$ , if $F^{'} (α) + 2$ is the area of the region bounded by $x = 0, y = 0, y = f (x)$ and $x = α$ , then $f (0)$ is

For $x \in (0, 5 π /2)$ , define $f (x) = \int_{0}^{x} t sin t d t$ . Then $f$ has

Let $f$ be a twice differentiable function on $R$ . If $f^{'} (0) = 4$ and $f (x) + \int_{0}^{x} (x - t) f^{'} (t) d t = (e^{2 x} + e^{- 2 x}) cos 2 x + \frac{2}{a} x$ , then $(2 a + 1)^{5} a^{2}$ is equal to

$lim_{x \to \frac{π}{2}} \frac{\int _{x^{3}}^{(π /2)^{3}} ( s i n ( 2 t ^{1/3} ) + c o s ( t ^{1/3} )) d t}{( x - \frac{π}{2} ) ^{2}}$ is equal to

If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is

$lim_{x \to \frac{π}{2}} (\frac{1}{( x - \frac{π}{2} ) ^{2}} \int_{x^{3}}^{(\frac{π}{2})^{3}} cos (t^{\frac{1}{3}}) d t)$ is equal to

Let $A_{n}$ be the area bounded by the curve $y = (tan x)^{n}$ and the lines $x = 0, y = 0$ and $x = \frac{π}{4}$ . Prove that for $n > 2, A_{n} + A_{n - 2} = \frac{1}{n - 1}$ and deduce $\frac{1}{2 n + 2} < A_{n} < \frac{1}{2 n - 2}$ .

Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0$ . Then $f (\frac{π ^{2}}{4})$ is equal to

Select Answer:

Visualized Solution

Visualizing the Curve .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=∫0x​∣t∣dt

The Newton-Leibniz Rule

Differentiating the Integral

Setting the Slope Condition

Finding the First Point of Tangency

Finding the Second Point of Tangency

Equation of the First Tangent

Equation of the Second Tangent

Calculating the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-intercepts

Final Conclusion

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Power of Newton-Leibniz

Finding the Points of Tangency

The Final Intercepts

Similar Questions

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Visualizing the Curve $y = \int_{0}^{x} ∣ t ∣ d t$

Calculating the $x$ -intercepts

The total number of distinct $x \in [0, 1]$ for which $\int_{0}^{x} \frac{t ^{2}}{1 + t ^{4}} d t = 2 x - 1$ is

Let $f (x) = \int_{1}^{x} 2 - t^{2} d t$ . Then the real roots of the equation $x^{2} - f^{'} (x) = 0$ are

For $x \in (0, 5 π /2)$ , define $f (x) = \int_{0}^{x} t sin t d t$ . Then $f$ has

Let $f$ be a twice differentiable function on $R$ . If $f^{'} (0) = 4$ and $f (x) + \int_{0}^{x} (x - t) f^{'} (t) d t = (e^{2 x} + e^{- 2 x}) cos 2 x + \frac{2}{a} x$ , then $(2 a + 1)^{5} a^{2}$ is equal to

$lim_{x \to \frac{π}{2}} \frac{\int _{x^{3}}^{(π /2)^{3}} ( s i n ( 2 t ^{1/3} ) + c o s ( t ^{1/3} )) d t}{( x - \frac{π}{2} ) ^{2}}$ is equal to

If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is

$lim_{x \to \frac{π}{2}} (\frac{1}{( x - \frac{π}{2} ) ^{2}} \int_{x^{3}}^{(\frac{π}{2})^{3}} cos (t^{\frac{1}{3}}) d t)$ is equal to

Let $A_{n}$ be the area bounded by the curve $y = (tan x)^{n}$ and the lines $x = 0, y = 0$ and $x = \frac{π}{4}$ . Prove that for $n > 2, A_{n} + A_{n - 2} = \frac{1}{n - 1}$ and deduce $\frac{1}{2 n + 2} < A_{n} < \frac{1}{2 n - 2}$ .

Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0$ . Then $f (\frac{π ^{2}}{4})$ is equal to