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Integral Calculus

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Question : 27 of 73

Marks: +1, -0

\int\limits_{0}^{4}\sqrt{3x+4}\,dx

is equal to

$\frac{9}{112}$
$\frac{112}{9}$
$\frac{11}{9}$
None

Solution:

Concept:

Evaluate the definite integral using substitution and power rule.

Explanation:

Let

u = 3x + 4

, then

du = 3\,dx

, so

dx = \frac{du}{3}

.

When

x = 0

,

u = 4

; when

x = 4

,

u = 16

.

The integral becomes:

\int_{0}^{4} \sqrt{3x+4}\,dx = \frac{1}{3} \int_{4}^{16} \sqrt{u}\,du = \frac{1}{3} \int_{4}^{16} u^{1/2}\,du

Integrate:

\frac{1}{3} \cdot \frac{2}{3} u^{3/2} \big|_{4}^{16} = \frac{2}{9} \left( 16^{3/2} - 4^{3/2} \right)

.

16^{3/2} = (16^{1/2})^3 = 4^3 = 64

,

4^{3/2} = (4^{1/2})^3 = 2^3 = 8

.

Thus,

\frac{2}{9}(64 - 8) = \frac{2}{9} \cdot 56 = \frac{112}{9}

.

Answer:

\frac{112}{9}

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