Definite IntegrationmediumFree

Range of f(x)=∫₁ˣ|t|dt on [−1/2, 1/2] | JEE

JEE Maths question with a full step-by-step solution.

Question
The range of the function f(x)=∫1x∣t∣ dtf(x)=\displaystyle\int_{1}^{x}|t|\,dt, x∈[−12,12]x\in\left[-\dfrac12,\dfrac12\right], is
A[38,58]\left[\dfrac38,\dfrac58\right]
B[−58,38]\left[-\dfrac58,\dfrac38\right]
C[−38,58]\left[-\dfrac38,\dfrac58\right]
D[−58,−38]\left[-\dfrac58,-\dfrac38\right]correct
Solution
Step 1: Differentiate using the Fundamental Theorem of Calculus: f′(x)=∣x∣f'(x)=|x|. On [−12,12]\left[-\dfrac12,\dfrac12\right], ∣x∣≥0|x|\ge 0, so ff is non-decreasing (strictly increasing away from 00). Therefore the range is [f(−12), f(12)]\big[f(-\tfrac12),\,f(\tfrac12)\big]. Step 2: Compute f ⁣(12)=∫11/2∣t∣ dt=−∫1/21t dt=−[t22]1/21=−(12−18)=−38.f\!\left(\dfrac12\right)=\displaystyle\int_{1}^{1/2}|t|\,dt=-\int_{1/2}^{1}t\,dt=-\left[\dfrac{t^{2}}{2}\right]_{1/2}^{1}=-\left(\dfrac12-\dfrac18\right)=-\dfrac38. Step 3: Compute f ⁣(−12)=∫1−1/2∣t∣ dt=−∫−1/21∣t∣ dtf\!\left(-\dfrac12\right)=\displaystyle\int_{1}^{-1/2}|t|\,dt=-\int_{-1/2}^{1}|t|\,dt. Split at 00:
∫−1/21∣t∣ dt=∫−1/20(−t) dt+∫01t dt=18+12=58,\int_{-1/2}^{1}|t|\,dt=\int_{-1/2}^{0}(-t)\,dt+\int_{0}^{1}t\,dt=\dfrac18+\dfrac12=\dfrac58,
 so f ⁣(−12)=−58.f\!\left(-\dfrac12\right)=-\dfrac58. Step 4: Since ff increases, the range is [f(−12),f(12)]=[−58,−38].\left[f(-\tfrac12),f(\tfrac12)\right]=\left[-\dfrac58,-\dfrac38\right]. Correct answer: (4)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Definite Integration · easy
∫0π/2sin⁡x1+cos⁡x+sin⁡x dx\displaystyle\int_{0}^{\pi/2}\dfrac{\sin x}{1+\cos x+\sin x}\,dx equals
Definite Integration · easy
If f(x)=∫1/xxcos⁡t2 dtf(x)=\displaystyle\int_{1/x}^{\sqrt{x}}\cos t^{2}\,dt for x>0x>0, then df(x)dx\dfrac{df(x)}{dx} is
Definite Integration · easy
If ∫0π/2sin⁡10x cos⁡10x dx=2−n∫0π/2(sin⁡10x+cos⁡10x) dx\displaystyle\int_{0}^{\pi/2}\sin^{10}x\,\cos^{10}x\,dx=2^{-n}\displaystyle\int_{0}^{\pi/2}\big(\sin^{10}x+\cos^{10}x\big)\,dx, n∈Nn\in\mathbb{N}, then n=n=
Definite Integration · easy
If ∫sin⁡x1t2f(t) dt=1−sin⁡x\displaystyle\int_{\sin x}^{1}t^{2}f(t)\,dt=1-\sin x for all x∈(0,π2]x\in\left(0,\dfrac{\pi}{2}\right], then the value of f ⁣(13)f\!\left(\dfrac{1}{\sqrt3}\right) is

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath