If - x-0x2-tdt- then d-dx is-

Φ (x)=∫_0^x^2√(t) dt=2/3 | t^3/2 |_0^x^2 ϕ (x)=2/3x^3 dϕ/dx=2x^2 Alternative Solution ϕ (x)=∫_0^x^2√(t) dt Apply Newton Leibnitz theorem dϕ/dx=0+2x √(x^2) 0=2 ...

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Byju's Answer

Standard XII

Mathematics

Parametric Differentiation

If ϕ x=∫0x2√t...

Question

If $ϕ (x) = \int_{0}^{x^{2}} \sqrt{t} d t .$ then $\frac{d ϕ}{d x}$ is:

2 x^{2}

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\sqrt{x}

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Solution

The correct option is A $2 x^{2}$
$ϕ (x) = \int_{0}^{x^{2}} \sqrt{t} d t = \frac{2}{3} {∣ ∣ t^{3 / 2} ∣ ∣}_{0}^{3 / 2}$
$ϕ (x) = \frac{2}{3} x^{3}$
$\frac{d ϕ}{d x} = 2 x^{2}$

Alternative Solution

$ϕ (x) = \int_{0}^{x^{2}} \sqrt{t} d t$
Apply Newton Leibnitz theorem
$\frac{d ϕ}{d x} = 0 + 2 x \sqrt{x^{2}} - 0 = 2 x^{2}$

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If ϕ(x)=∫x20√tdt. then dϕdx is:

The correct option is A 2x2ϕ(x)=∫x20√tdt=23∣∣t3/2∣∣x20 ϕ(x)=23x3 dϕdx=2x2 Alternative Solution ϕ(x)=∫x20√tdt Apply Newton Leibnitz theorem dϕdx=0+2x√x2−0=2x2

If $ϕ (x) = \int_{0}^{x^{2}} \sqrt{t} d t .$ then $\frac{d ϕ}{d x}$ is: