Evaluate- -0-3-x3 -1- dx

The correct option is C 94 √(3) Consider, I= ∫_0^√(3)[x^3 1] dx ⇒ I= [x^44 x]^√(3)_0 I=94 √(3) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

Evaluate: ∫...

Question

Evaluate: $\int_{0}^{\sqrt{3}} [x^{3} - 1] d x$

\frac{1}{4} - \sqrt{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{4} - \sqrt{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{9}{4} - \sqrt{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{9}{4} - \sqrt{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{9}{4} - \sqrt{3}$
Consider, $I = \int_{0}^{\sqrt{3}} [x^{3} - 1] d x$

$\Rightarrow I = {[\frac{x^{4}}{4} - x]}_{0}^{\sqrt{3}}$

$I = \frac{9}{4} - \sqrt{3}$

Suggest Corrections

Similar questions

Evaluate : $(i) \frac{5}{9} + \frac{- 7}{6} (i i) 4 + \frac{3}{- 5} (i i i) \frac{1}{- 15} + \frac{5}{- 12} (i v) \frac{5}{9} + \frac{3}{- 4} (v) \frac{- 8}{9} + \frac{- 5}{12} (v i) 0 + \frac{- 2}{7} (v i i) \frac{5}{- 11} + 0 (v i i i) 2 + \frac{- 3}{5} (i x) \frac{4}{- 9} + 1$

Evaluate: $30 - [26 - {15 + (8 - ¯ ¯¯¯¯¯¯¯¯¯¯ ¯ 6 - 3)}] \times \frac{8}{15} \div \frac{4}{9} \times \frac{3}{4} - \frac{1}{2} + 1$

Evaluate : $(3 a - \frac{1}{2}) (3 a - \frac{1}{2})$

Q. Evaluate :

(3 a - \frac{1}{2}) (3 a - \frac{1}{2})

Q. Evaluate :

4 (2 x - 3)^{2} - 9 (y + 1)^{2}

Join BYJU'S Learning Program

Evaluate: ∫√30[x3−1]dx

The correct option is C 94−√3Consider, I=∫√30[x3−1]dx⇒I=[x44−x]√30I=94−√3

Evaluate: $\int_{0}^{\sqrt{3}} [x^{3} - 1] d x$

The correct option is C $\frac{9}{4} - \sqrt{3}$
Consider, $I = \int_{0}^{\sqrt{3}} [x^{3} - 1] d x$

$\Rightarrow I = {[\frac{x^{4}}{4} - x]}_{0}^{\sqrt{3}}$

$I = \frac{9}{4} - \sqrt{3}$