Question

If $\int_{0}^{a}\sqrt{x}dx = \frac{4a}{3}$, then $\int_{a}^{a+1}x\,dx$ is:

• A. $\frac{3}{2}$
• B. $\frac{9}{2}$
• C. $\frac{5}{2}$
• D. $\frac{7}{2}$

Hint:

Always expand binomial terms before substituting values in definite integrals.

Answer 1

The Correct Option is C

Concept: Use definite integration formula: \[ \int x\,dx=\frac{x^2}{2} \]

Step 1: {Apply limits.}
\[ \int_{a}^{a+1}x\,dx=\left[\frac{x^2}{2}\right]_{a}^{a+1} \]

Step 2: {Substitute upper limit.}
\[ \frac{(a+1)^2}{2} \]

Step 3: {Substitute lower limit.}
\[ \frac{a^2}{2} \]

Step 4: {Subtract.}
\[ \frac{(a+1)^2-a^2}{2} \]

Step 5: {Expand numerator.}
\[ \frac{a^2+2a+1-a^2}{2} \]

Step 6: {Simplify.}
\[ \frac{2a+1}{2} \]

Step 7: {Using given condition, solve for $a=2$.}

Step 8: {Final value.}
\[ \frac{5}{2} \]