If $\displaystyle\int_0^x f(t)\,dt = x+\int_x^1 tf(t)\,dt$, then the value of $f(1)$ is (1) $\dfrac12$ (2) $2$ (3) $1$ (4) $\dfrac34$

If ∫₀x f(t)dt = x+∫ₓ¹ tf(t)dt, then the value of f(1) is (1) dfrac12 (2) 2 (3) 1 (4) dfrac34

Exercise 9.10 · Q20

\displaystyle\int_0^x f(t)\,dt = x+\int_x^1 tf(t)\,dt

(1)

\dfrac12

(2)

2

(3)

1

(4)

\dfrac34

Puducherry TnboardTextbookSubjectiveImportance★★★★★

52% · 50/96 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Differentiating the given integral equation term by term (variable-upper-limit and variable-lower-limit integrals both reduce via the Fundamental Theorem) produces a simple algebraic equation for $f(x)$ , from which $f(1)$ follows directly.

Step 1. Write down the given equation.

\int_0^x f(t)\,dt = x+\int_x^1 tf(t)\,dt.

Step 2. Differentiate the left side w.r.t. $x$ . By the First Fundamental Theorem (Theorem 9.1),

\frac{d}{dx}\int_0^x f(t)\,dt=f(x).

Step 3. Differentiate the right side term by term. $\dfrac{d}{dx}(x)=1$ . For $\displaystyle\int_x^1 tf(t)\,dt=-\int_1^x tf(t)\,dt$ , applying the First Fundamental Theorem to $\int_1^x tf(t)dt$ (derivative $=xf(x)$ ) and the outer minus sign gives

$\frac{d}{dx}\int_x^1 tf(t)\,dt=-xf(x).$ …

Start your 14-day free trial to unlock the full solution →

←Previous question · 9.10Q19 — The value of ∫₀a (√(a²-x²))³dx is (1) (π a³)/(16) (2) (3π a⁴)/(16) (3)…→Next section · 9.11Summary

Q.If $\displaystyle\int_0^x f(t)\,dt = x+\int_x^1 tf(t)\,dt$ , then the value of $f(1)$ is

Concept understanding — Fundamental Theorem of Integral Calculus

Q.If ∫0xf(t) dt=x+∫x1tf(t) dt\displaystyle\int_0^x f(t)\,dt = x+\int_x^1 tf(t)\,dt∫0x​f(t)dt=x+∫x1​tf(t)dt, then the value of f(1)f(1)f(1) is

🎓Unlock everything free for 14 days

Q.If $\displaystyle\int_0^x f(t)\,dt = x+\int_x^1 tf(t)\,dt$ , then the value of $f(1)$ is

Unlock everything free for 14 days