Evaluate ∫ 1 3 x 2 + 3 x + e x d x as the limit of the sum.

Given, ∫ 1 3 x 2 + 3 x + e x d x Integral as limit of sum is given by, ∫ a b f x d x = lim h → 0 h f a + f a + h + f a + 2 h + ⋯ + f a + n - 1 h , where h = b - a n . Here, a = 1 and b = 3 ∴ h = 3 - 1 n ⇒ n h = 2 and f x = x 2 + 3 x + e x Now, ∫ 1 3 x 2 + 3 x + e x d x = lim h → 0 h f ( 1 ) + f ( 1 + h ) + f ( 1 + 2 h ) + … + f ( 1 + ( n - 1 ) h ) = lim h → 0 h 4 + e + ( 1 + h ) 2 + 3 ( 1 + h ) + e 1 + h + ( 1 + 2 h ) 2 + 3 ( 1 + 2 h ) + e 1 + 2 h + … + ( 1 + ( n - 1 ) h ) 2 + 3 ( 1 + ( n - 1 ) h ) + e 1 + ( n - 1 ) h = lim h → 0 h 4 n + e + h 2 1 + 2 2 + 3 2 + … ( n - 1 ) 2 + 2 h ( 1 + 2 + 3 + ( n - 1 ) ) + 3 h ( 1 + 2 + 3 … ( n - 1 ) ) + e 1 + h + e 1 + 2 h + … e 1 + ( n - 1 ) h = lim h → 0 h · 4 n + e + h 2 ( n - 1 ) ( n - 2 ) ( 2 n - 3 ) 6 + 2 h n ( n - 1 ) 2 + 3 h n ( n - 1 ) 2 + e e h e h . ( n - 1 ) - 1 e h - 1 = 8 + 8 3 + 4 + 6 + e 3 - e [Since n h = 2 ] = 62 3 + e 3 - e Hence, ∫ 1 3 x 2 + 3 x + e x d x = 62 3 + e 3 - e .

Evaluate ∫ 1 3 | 2 x - 1 | d x .

Given that, the integral is ∫ 1 3 | 2 x - 1 | d x ∫ 1 3 | 2 x - 1 | d x = ∫ 1 3 ( 2 x - 1 ) d x = 1 4 ( 2 x - 1 ) 2 1 3 = 6

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Important Questions on Integrals

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Mathematics>Integral Calculus>Integrals>Definite Integration

Let

F : R \to R

be a thrice differentiable function. Suppose that

F (1) = 0, F (3) = - 4

and

F^{'} (x) < 0

for all

x \in (\frac{1}{2}, 3) .

Let

f (x) = x F (x)

for all

x \in R

. If

\int_{1}^{3} x^{2} F^{'} (x) d x = - 12

and

\int_{1}^{3} x^{3} F^{''} (x) d x = 40,

then the correct expression(s) is(are)

EASY

Mathematics>Integral Calculus>Integrals>Definite Integration

The value of

\int_{0}^{2} \frac{d x}{x^{2} - 2 x + 2}

is equal to

HARD

Mathematics>Integral Calculus>Integrals>Definite Integration

\int_{1}^{2} \frac{d x}{{(x^{2} - 2 x + 4)}^{\frac{3}{2}}} = \frac{k}{k + 5}

, then

k

is equal to

MEDIUM

Mathematics>Integral Calculus>Integrals>Definite Integration

\int_{0}^{\frac{π}{2}} \frac{c o t x}{c o t x + cosec x} d x = m (π + n),

then

m n

is equal to

EASY

Mathematics>Integral Calculus>Integrals>Definite Integration

\sin^{- 1} x = y

, then

HARD

Mathematics>Integral Calculus>Integrals>Definite Integration

Let

f^{'} (x) = \frac{192 x^{3}}{2 + \sin^{4} π x}

for all

x \in R

with

f (\frac{1}{2}) = 0 .

m \leq \int_{\frac{1}{2}}^{1} f (x) d x \leq M,

then the possible values of

m

and

M

are

EASY

Mathematics>Integral Calculus>Integrals>Definite Integration

The integral

\int_{\frac{π}{12}}^{\frac{π}{4}} \frac{8 \cos 2 x}{{(\tan x + \cot x)}^{3}} d x

equals

HARD

Mathematics>Integral Calculus>Integrals>Definite Integration

Evaluate

\int_{1}^{3} (x^{2} + 3 x + e^{x}) d x

as the limit of the sum.

EASY

Mathematics>Integral Calculus>Integrals>Definite Integration

What is the correct IUPAC name of the given compound?

Question Image

MEDIUM

Mathematics>Integral Calculus>Integrals>Definite Integration

\int_{0}^{π / 3} \frac{\tan θ}{\sqrt{2 k \sec θ}} d θ = 1 - \frac{1}{\sqrt{2}}, (k > 0)

, then the value of

k

MEDIUM

Mathematics>Integral Calculus>Integrals>Definite Integration

The integral

\int_{\frac{π}{6}}^{\frac{π}{3}} {s e c}^{\frac{2}{3}} x \cdot c o s e c^{\frac{4}{3}} x d x

is equal to

MEDIUM

Mathematics>Integral Calculus>Integrals>Definite Integration

α = {c o s}^{- 1} (\frac{3}{5})

β = {t a n}^{- 1} (\frac{1}{3})

, where

0 < α, β < \frac{π}{2},

then

α - β

is equal to

HARD

Mathematics>Integral Calculus>Integrals>Definite Integration

\underset{n \to \infty}{l i m} (\frac{{(n + 1)}^{1 / 3}}{n^{4 / 3}} + \frac{{(n + 2)}^{1 / 3}}{n^{4 / 3}} + . . . . . + \frac{{(2 n)}^{1 / 3}}{n^{4 / 3}})

is equal to

HARD

Mathematics>Integral Calculus>Integrals>Definite Integration

Let

f (x) = 7 \tan^{8} x + 7 \tan^{6} x - 3 \tan^{4} x - 3 \tan^{2} x

for all

x \in (- \frac{π}{2}, \frac{π}{2}) .

Then the correct expression(s) is(are)

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Mathematics>Integral Calculus>Integrals>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 $\int_{0}^{π} (f^{'} (x) + F (x)) \cos x d x = 2$ , then the value of $f (0)$ is equal to