Evaluate the integral. $\int _ { 0 } ^ { \sqrt { 2 } } 2 x \sqrt { x ^ { 2 } + 9 } \mathrm {~d} x$
A) $\frac { 2 } { 3 } \left( 11 ^ { \frac { 3 } { 2 } } - 27 \right)$
B) $- \frac { 2 \cdot 11 ^ { \frac { 2 } { 3 } } } { 3 }$
C) $\frac { 2 \cdot 11 ^ { \frac { 3 } { 2 } } } { 3 }$
D) $\frac { 2 } { 3 } \left( 27 - 11 ^ { \frac { 3 } { 2 } } \right)$
E) $\frac { 2 } { 3 } \left( 11 ^ { \frac { 2 } { 3 } } - 27 \right)$

Question

Evaluate the integral.   $\int _ { 0 } ^ { \sqrt { 2 } } 2 x \sqrt { x ^ { 2 } + 9 } \mathrm {~d} x$  
A) $\frac { 2 } { 3 } \left( 11 ^ { \frac { 3 } { 2 } } - 27 \right)$
B) $- \frac { 2 \cdot 11 ^ { \frac { 2 } { 3 } } } { 3 }$
C) $\frac { 2 \cdot 11 ^ { \frac { 3 } { 2 } } } { 3 }$
D) $\frac { 2 } { 3 } \left( 27 - 11 ^ { \frac { 3 } { 2 } } \right)$
E) $\frac { 2 } { 3 } \left( 11 ^ { \frac { 2 } { 3 } } - 27 \right)$

Quizplus · Accepted Answer

To solve the integral $\int _ { 0 } ^ { \sqrt { 2 } } 2 x \sqrt { x ^ { 2 } + 9 } \mathrm {~d} x$, we can use a substitution method. Let $u = x^2 + 9$, then $du = 2x dx$. The integral becomes $\int \sqrt{u} \, du$, which evaluates to $\frac{2}{3}u^{3/2}$. Substituting back for $u$ and evaluating from $0$ to $\sqrt{2}$ gives $\frac{2}{3}(11^{3/2} - 27)$.

Evaluate the Integral ∫022xx2+9 dx\int _ { 0 } ^ { \sqrt { 2 } } 2 x \sqrt { x ^ { 2 } + 9 } \mathrm {~d} x∫02​​2xx2+9​ dx

Evaluate the Integral $\int _ { 0 } ^ { \sqrt { 2 } } 2 x \sqrt { x ^ { 2 } + 9 } \mathrm {~d} x$