The correct option is B 15+e^82 Now, ∫^4_0 (x + e^2x)dx =[x^22]_0^4+[e^2x2]_0^4 =162+e^8 12 =15+e^82 . ...

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Standard XII

Mathematics

Property 1

∫40 x + e2xdx

Question

$\int_{0}^{4} (x + e^{2 x}) d x$

\frac{15}{2}

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\frac{15 + e^{8}}{2}

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\frac{15 - e^{8}}{2}

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\frac{e^{8}}{2}

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Solution

The correct option is B $\frac{15 + e^{8}}{2}$
Now,
$\int_{0}^{4} (x + e^{2 x}) d x$
$= {[\frac{x^{2}}{2}]}_{0}^{4} + {[\frac{e^{2 x}}{2}]}_{0}^{4}$
$= \frac{16}{2} + \frac{e^{8} - 1}{2}$
$= \frac{15 + e^{8}}{2}$ .

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∫40(x+e2x)dx

The correct option is B 15+e82Now,∫40(x+e2x)dx=[x22]40+[e2x2]40=162+e8−12=15+e82.

$\int_{0}^{4} (x + e^{2 x}) d x$

The correct option is B $\frac{15 + e^{8}}{2}$
Now,
$\int_{0}^{4} (x + e^{2 x}) d x$
$= {[\frac{x^{2}}{2}]}_{0}^{4} + {[\frac{e^{2 x}}{2}]}_{0}^{4}$
$= \frac{16}{2} + \frac{e^{8} - 1}{2}$
$= \frac{15 + e^{8}}{2}$ .