Evaluate 2^{x} x² d x

Correct Answer is : 2xx2log⁡2− x2x+1(log⁡2)2+ 2x+1(log⁡2)3+c\;{2^{x} x²}{ 2} - \;{x 2^{x+1}}{( 2)²} + \;{2^{x+1}}{( 2)³} + clog22xx2−(log2)2x2x+1+(log2)32x+1+c

\;{2^{x} x²}{ 2} - \;{x 2^{x+1}}{( 2)²} + \;{2^{x+1}}{( 2)³} + c

\;{2^{x} x³}{3} - \;{x² 2^{x+1}}{( 2)²} + \;{2^{x+1}}{( 3)²} + c

\;{2^{x} · x²}{3} - \;{x³ · 2^{x}}{3} - \;{2^{x+1}}{( 2)³} + c

Evaluate 2^{x} x² d x

Correct Answer is : 2xx2log⁡2− x2x+1(log⁡2)2+ 2x+1(log⁡2)3+c\;{2^{x} x²}{ 2} - \;{x 2^{x+1}}{( 2)²} + \;{2^{x+1}}{( 2)³} + clog22xx2−(log2)2x2x+1+(log2)32x+1+c

Test Index

Integral Calculus

Question : 31 of 73

Marks: +1, -0

Evaluate

\int 2^{x} x^{2} d x

$\;\frac{2^{x} x^{2}}{\log 2} - \;\frac{x 2^{x+1}}{(\log 2)^{2}} + \;\frac{2^{x+1}}{(\log 2)^{3}} + c$
$\;\frac{2^{x} x^{3}}{3} - \;\frac{x^{2} 2^{x+1}}{(\log 2)^{2}} + \;\frac{2^{x+1}}{(\log 3)^{2}} + c$
$\;\frac{2^{x} \cdot x^{2}}{3} - \;\frac{x^{3} \cdot 2^{x}}{3} - \;\frac{2^{x+1}}{(\log 2)^{3}} + c$
None

Validate

Solution:

Concept:

Use integration by parts repeatedly to reduce the power of

x

Explanation:

Let

I = \int 2^{x} x^{2} dx

Set

u = x^{2}

dv = 2^{x} dx

. Then

du = 2x dx

v = \frac{2^{x}}{\ln 2}

Integration by parts:

I = x^{2} \cdot \frac{2^{x}}{\ln 2} - \int \frac{2^{x}}{\ln 2} \cdot 2x dx = \frac{x^{2}2^{x}}{\ln 2} - \frac{2}{\ln 2} \int x 2^{x} dx

Now evaluate

J = \int x 2^{x} dx

. Set

u = x

dv = 2^{x} dx

, so

du = dx

v = \frac{2^{x}}{\ln 2}

Then

J = x \cdot \frac{2^{x}}{\ln 2} - \int \frac{2^{x}}{\ln 2} dx = \frac{x 2^{x}}{\ln 2} - \frac{1}{\ln 2} \cdot \frac{2^{x}}{\ln 2} = \frac{x 2^{x}}{\ln 2} - \frac{2^{x}}{(\ln 2)^{2}}

Substitute back:

I = \frac{x^{2}2^{x}}{\ln 2} - \frac{2}{\ln 2} \left( \frac{x 2^{x}}{\ln 2} - \frac{2^{x}}{(\ln 2)^{2}} \right)

Simplify:

I = \frac{x^{2}2^{x}}{\ln 2} - \frac{2x 2^{x}}{(\ln 2)^{2}} + \frac{2 \cdot 2^{x}}{(\ln 2)^{3}} = \frac{2^{x} x^{2}}{\ln 2} - \frac{x 2^{x+1}}{(\ln 2)^{2}} + \frac{2^{x+1}}{(\ln 2)^{3}} + c

This matches option A.

Answer:

\frac{2^{x} x^{2}}{\log 2} - \frac{x 2^{x+1}}{(\log 2)^{2}} + \frac{2^{x+1}}{(\log 2)^{3}} + c

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