Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that its derivative $f'$ is continuous and $f(\pi) = -6$. If $F: [0, \pi] \to \mathbb{R}$ is defined by $F(x) = \int_0^x f(t) dt$, and if $\int_0^\pi (f'(x) + F(x)) \cos x dx = 2$, then the value of $f(0)$ is ____.

### Analyzing the Setup We are given a differentiable function $f: \mathbb{R} \to \mathbb{R}$ with the boundary condition $f(\pi) = -6$. The problem involves an integral equation that we must decompose to isolate the unknown value $f(0)$. We split the integral into two distinct parts: $$I_1 = \int_0^\pi f'(x) \cos x \, dx$$ $$I_2 = \int_0^\pi F(x) \cos x \, dx$$ ### Evaluating the First Integral For $I_1$, we apply the method of **Integration by Parts**. Let $u = \cos x$ and $dv = f'(x) \, dx$. This yields: $$I_1 = [f(x) \cos x]_0^\pi - \int_0^\pi f(x) (-\sin x) \, dx$$ Simplifying the expression, we obtain: $$I_1 = [f(x) \cos x]_0^\pi + \int_0^\pi f(x) \sin x \, dx$$ Evaluating the boundary terms at $\pi$ and $0$: $$I_1 = f(\pi) \cos(\pi) - f(0) \cos(0) + \int_0^\pi f(x) \sin x \, dx$$ $$I_1 = -f(\pi) - f(0) + \int_0^\pi f(x) \sin x \, dx$$ Given $f(\pi) = -6$, the expression becomes: $$I_1 = 6 - f(0) + \int_0^\pi f(x) \sin x \, dx$$ ### Evaluating the Second Integral For $I_2$, we again use **Integration by Parts**. Let $u = F(x)$ and $dv = \cos x \, dx$. This gives: $$I_2 = [F(x) \sin x]_0^\pi - \int_0^\pi F'(x) \sin x \, dx$$ Since $F'(x) = f(x)$ and $\sin(\pi) = \sin(0) = 0$, the boundary term vanishes: $$I_2 = 0 - \int_0^\pi f(x) \sin x \, dx$$ ### Final Calculation When we combine $I_1$ and $I_2$, the integral terms $\int_0^\pi f(x) \sin x \, dx$ and $-\int_0^\pi f(x) \sin x \, dx$ cancel each other out perfectly. We are left with the simplified equation: $$6 - f(0) = 2$$ Solving for the unknown, we find: **$f(0) = 4$**

JEE Advanced 2020

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let $f : R \to R$ be a differentiable function such that its derivative $f^{'}$ is continuous and $f (π) = - 6$ . If $F : [0, π] \to R$ is defined by $F (x) = \int_{0}^{x} f (t) d t$ , and if $\int_{0}^{π} (f^{'} (x) + F (x)) cos x d x = 2$ , then the value of $f (0)$ is ____.

Q: How do I know which function to pick as $u$ and which as $dv$ in Integration by Parts?

Think of the goal: you want to reduce the complexity. For $I_1$, we pick $dv = f'(x)dx$ because we know its integral is $f(x)$. For $I_2$, we pick $u = F(x)$ because we know its derivative is $f(x)$. Always choose $dv$ such that it is easily integrable, and $u$ such that its derivative simplifies the expression!

Enter Numerical Value:

Visualized Solution

Splitting the Integral

Given: $\int_{0}^{π} (f^{'} (x) + F (x)) cos x d x = 2$
Let's split this into two separate integrals:
$I_{1} = \int_{0}^{π} f^{'} (x) cos x d x$
$I_{2} = \int_{0}^{π} F (x) cos x d x$
So, $I_{1} + I_{2} = 2$

Integration by Parts for $I_{1}$

Focus on $I_{1} = \int_{0}^{π} f^{'} (x) cos x d x$
We will use Integration by Parts: $\int u d v = uv - \int v d u$
Choose $u = cos x$ (to differentiate)
Choose $d v = f^{'} (x) d x$ (to integrate)

Applying the Formula to $I_{1}$

Differentiating $u$ : $d u = - sin x d x$
Integrating $d v$ : $v = f (x)$
Applying the formula to $I_{1}$ :
$I_{1} = [f (x) cos x]_{0}^{π} - \int_{0}^{π} f (x) (- sin x) d x$

Evaluating Limits for $I_{1}$

Let's evaluate the boundary term: $[f (x) cos x]_{0}^{π}$
Upper limit: $f (π) cos π = f (π) (- 1) = - f (π)$
Lower limit: $f (0) cos 0 = f (0) (1) = f (0)$
Boundary term becomes: $- f (π) - f (0)$

Simplifying $I_{1}$

Substitute the boundary term back into $I_{1}$ :
$I_{1} = - f (π) - f (0) + \int_{0}^{π} f (x) sin x d x$
We are given $f (π) = - 6$
$I_{1} = - (- 6) - f (0) + \int_{0}^{π} f (x) sin x d x$
$I_{1} = 6 - f (0) + \int_{0}^{π} f (x) sin x d x$

Integration by Parts for $I_{2}$

Now focus on $I_{2} = \int_{0}^{π} F (x) cos x d x$
Again, use Integration by Parts.
Choose $u = F (x)$ (to differentiate)
Choose $d v = cos x d x$ (to integrate)

Applying the Formula to $I_{2}$

Differentiating $u$ : $d u = F^{'} (x) d x$
Integrating $d v$ : $v = sin x$
Applying the formula to $I_{2}$ :
$I_{2} = [F (x) sin x]_{0}^{π} - \int_{0}^{π} F^{'} (x) sin x d x$

Evaluating Limits for $I_{2}$

Let's evaluate the boundary term: $[F (x) sin x]_{0}^{π}$
Upper limit: $F (π) sin π = F (π) (0) = 0$
Lower limit: $F (0) sin 0 = F (0) (0) = 0$
The boundary term completely vanishes!

Fundamental Theorem of Calculus

We know $F (x) = \int_{0}^{x} f (t) d t$
By the Fundamental Theorem of Calculus:
$F^{'} (x) = f (x)$
Substitute this into our remaining integral for $I_{2}$ :
$I_{2} = 0 - \int_{0}^{π} f (x) sin x d x$

Combining $I_{1}$ and $I_{2}$

Recall our original equation: $I_{1} + I_{2} = 2$
Substitute the simplified forms of $I_{1}$ and $I_{2}$ :
$(6 - f (0) + \int_{0}^{π} f (x) sin x d x) + (- \int_{0}^{π} f (x) sin x d x) = 2$

Canceling the Integrals

Notice the integral terms: $+ \int_{0}^{π} f (x) sin x d x$ and $- \int_{0}^{π} f (x) sin x d x$
They perfectly cancel each other out!
We are left with a simple algebraic equation:
$6 - f (0) = 2$

Solving for $f (0)$

Rearrange the equation to solve for $f (0)$ :
$f (0) = 6 - 2$
$f (0) = 4$

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

We are given a differentiable function

f : R \to R

with the boundary condition

f (π) = - 6

. The problem involves an integral equation that we must decompose to isolate the unknown value

f (0)

We split the integral into two distinct parts:

I_{1} = \int_{0}^{π} f^{'} (x) cos x d x

I_{2} = \int_{0}^{π} F (x) cos x d x

Evaluating the First Integral

For

I_{1}

, we apply the method of Integration by Parts. Let

u = cos x

and

d v = f^{'} (x) d x

This yields:

I_{1} = [f (x) cos x]_{0}^{π} - \int_{0}^{π} f (x) (- sin x) d x

Simplifying the expression, we obtain:

I_{1} = [f (x) cos x]_{0}^{π} + \int_{0}^{π} f (x) sin x d x

Evaluating the boundary terms at

π

and

0

I_{1} = f (π) cos (π) - f (0) cos (0) + \int_{0}^{π} f (x) sin x d x

I_{1} = - f (π) - f (0) + \int_{0}^{π} f (x) sin x d x

Given

f (π) = - 6

, the expression becomes:

I_{1} = 6 - f (0) + \int_{0}^{π} f (x) sin x d x

Evaluating the Second Integral

For

I_{2}

, we again use Integration by Parts. Let

u = F (x)

and

d v = cos x d x

This gives:

I_{2} = [F (x) sin x]_{0}^{π} - \int_{0}^{π} F^{'} (x) sin x d x

Since

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

and

sin (π) = sin (0) = 0

, the boundary term vanishes:

I_{2} = 0 - \int_{0}^{π} f (x) sin x d x

Final Calculation

When we combine

I_{1}

and

I_{2}

, the integral terms

\int_{0}^{π} f (x) sin x d x

and

- \int_{0}^{π} f (x) sin x d x

cancel each other out perfectly.

We are left with the simplified equation:

6 - f (0) = 2

Solving for the unknown, we find: $f (0) = 4$

Common Traps & FAQs

Why do we split the integral? Can't we just integrate it as one piece?

Great question! In JEE Advanced problems, the structure is often the hint. If you try to integrate

(f^{'} (x) + F (x)) cos x

directly, you hit a wall because you don't know the explicit form of

f (x)

. By splitting, you isolate the components, allowing the Fundamental Theorem of Calculus to reveal the hidden symmetry between

f^{'} (x)

and

F (x)

How do I know which function to pick as

u

and which as

d v

in Integration by Parts?

Think of the goal: you want to reduce the complexity. For

I_{1}

, we pick

d v = f^{'} (x) d x

because we know its integral is

f (x)

. For

I_{2}

, we pick

u = F (x)

because we know its derivative is

f (x)

. Always choose

d v

such that it is easily integrable, and

u

such that its derivative simplifies the expression!

I ended up with an integral of

f (x) sin x

that I can't solve. Did I mess up?

Not at all! That is the 'trap' meant to test your confidence. In many of these problems, the integral you cannot solve is actually the one that cancels out. Trust the process—the math will resolve itself if you have performed the Integration by Parts correctly. The cancellation is the reward for your persistence.

Enter Numerical Value:

Visualized Solution

Splitting the Integral

Integration by Parts for $I_{1}$

Applying the Formula to $I_{1}$

Evaluating Limits for $I_{1}$

Simplifying $I_{1}$

Integration by Parts for $I_{2}$

Applying the Formula to $I_{2}$

Evaluating Limits for $I_{2}$

Fundamental Theorem of Calculus

Combining $I_{1}$ and $I_{2}$

Canceling the Integrals

Solving for $f (0)$

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Evaluating the First Integral

Evaluating the Second Integral

Final Calculation

Similar Questions

Let $f : R \to R$ be a differentiable function such that $f (0) = 0$ , $f^{'} (\frac{π}{2}) = 3$ and $f^{'} (0) = 1$ . If $g (x) = \int_{x}^{π /2} [f^{'} (t) cosec t - cot t cosec t f (t)] d t$ for $x \in (0, \frac{π}{2})$ , then $lim_{x \to 0} g (x) =$

Let $f (θ) = sin θ + \int_{- π /2}^{π /2} (sin θ + t cos θ) f (t) d t$ . Then the value of $\int_{0}^{π /2} f (θ) d θ$ is ______.

Let $f (x)$ be a differentiable function defined on $[0, 2]$ such that $f^{'} (x) = f^{'} (2 - x)$ for all $x \in (0, 2)$ , $f (0) = 1$ and $f (2) = e^{2}$ . Then the value of $\int_{0}^{2} f (x) d x$ is:

Let $f : R \to R$ be a twice differentiable function such that $f (2) = 1.$ If $F (x) = x f (x)$ for all $x \in R$ , $\int_{0}^{2} x F^{'} (x) d x = 6$ and $\int_{0}^{2} x^{2} F^{''} (x) d x = 40$ then $F^{'} (2) + \int_{0}^{2} F (x) d x$ is equal to:

Let $f (x)$ be a function satisfying $f (x) + f (π - x) = π^{2}, \forall x \in R$ . Then $\int_{0}^{π} f (x) sin x d x$ is equal to

If $f (x)$ satisfies the relation $f (x) = e^{x} + \int_{0}^{1} (y + x e^{x}) f (y) d y$ , then $e + f (0)$ is equal to .........

If $f (t) = \int_{0}^{π} \frac{2 x d x}{1 - c o s ^{2} t s i n ^{2} x}, 0 < t < π$ , then the value of $\int_{0}^{π /2} \frac{π ^{2} d t}{f ( t )}$ equals

The value of the integral $\frac{48}{π ^{4}} \int_{0}^{π} x^{2} (\frac{3 π x ^{2}}{2} - x^{3}) \frac{s i n x}{1 + c o s ^{2} x} d x$ is equal to ______.

The value of the integral $\int_{- π /4}^{π /4} \frac{x + \frac{π}{4}}{2 - c o s 2 x} d x$ is

Enter Numerical Value:

Visualized Solution

Splitting the Integral

Integration by Parts for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Applying the Formula to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Evaluating Limits for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Integration by Parts for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Applying the Formula to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Evaluating Limits for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Fundamental Theorem of Calculus

Combining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Canceling the Integrals

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0)

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Evaluating the First Integral

Evaluating the Second Integral

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a differentiable function defined on [0,2] such that f′(x)=f′(2−x) for all x∈(0,2), f(0)=1 and f(2)=e2. Then the value of ∫02​f(x)dx is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a function satisfying f(x)+f(π−x)=π2,∀x∈R. Then ∫0π​f(x)sinxdx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x) satisfies the relation f(x)=ex+∫01​(y+xex)f(y)dy, then e+f(0) is equal to .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(t)=∫0π​1−cos2tsin2x2xdx​,0<t<π, then the value of ∫0π/2​f(t)π2dt​ equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of the integral π448​∫0π​x2(23πx2​−x3)1+cos2xsinx​dx is equal to ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of the integral ∫−π/4π/4​2−cos2xx+4π​​dx is

Integration by Parts for $I_{1}$

Applying the Formula to $I_{1}$

Evaluating Limits for $I_{1}$

Simplifying $I_{1}$

Integration by Parts for $I_{2}$

Applying the Formula to $I_{2}$

Evaluating Limits for $I_{2}$

Combining $I_{1}$ and $I_{2}$

Solving for $f (0)$

Let $f (θ) = sin θ + \int_{- π /2}^{π /2} (sin θ + t cos θ) f (t) d t$ . Then the value of $\int_{0}^{π /2} f (θ) d θ$ is ______.

Let $f (x)$ be a differentiable function defined on $[0, 2]$ such that $f^{'} (x) = f^{'} (2 - x)$ for all $x \in (0, 2)$ , $f (0) = 1$ and $f (2) = e^{2}$ . Then the value of $\int_{0}^{2} f (x) d x$ is:

Let $f (x)$ be a function satisfying $f (x) + f (π - x) = π^{2}, \forall x \in R$ . Then $\int_{0}^{π} f (x) sin x d x$ is equal to

If $f (x)$ satisfies the relation $f (x) = e^{x} + \int_{0}^{1} (y + x e^{x}) f (y) d y$ , then $e + f (0)$ is equal to .........

If $f (t) = \int_{0}^{π} \frac{2 x d x}{1 - c o s ^{2} t s i n ^{2} x}, 0 < t < π$ , then the value of $\int_{0}^{π /2} \frac{π ^{2} d t}{f ( t )}$ equals

The value of the integral $\frac{48}{π ^{4}} \int_{0}^{π} x^{2} (\frac{3 π x ^{2}}{2} - x^{3}) \frac{s i n x}{1 + c o s ^{2} x} d x$ is equal to ______.

The value of the integral $\int_{- π /4}^{π /4} \frac{x + \frac{π}{4}}{2 - c o s 2 x} d x$ is