AP Calculus AB and BC

Integral of sqrt x: Answer, Steps, and Mistakes

Q: What is the integral of sqrt x?

\int \sqrt{x}\,dx = \frac{2}{3}x^{3/2} + C. Rewrite \sqrt{x} as x^{1/2}, add 1 to the exponent to get \frac{3}{2}, then divide by \frac{3}{2}, which multiplies by \frac{2}{3}.

Q: What is the integral of sqrt x from 0 to 4?

It is \frac{16}{3}. Evaluate \frac{2}{3}x^{3/2} at the endpoints: 4^{3/2} = 8 because \sqrt{4} = 2 and 2^3 = 8, so you get \frac{2}{3}(8) - 0 = \frac{16}{3}.

Q: How do you integrate sqrt(ax + b)?

Use u-substitution, not the bare power rule. With u = ax + b you have du = a\,dx, so \int \sqrt{ax+b}\,dx = \frac{2}{3a}(ax+b)^{3/2} + C. For example \int \sqrt{3x+1}\,dx = \frac{2}{9}(3x+1)^{3/2} + C.

Q: What is the domain of the antiderivative?

Both \sqrt{x} and \frac{2}{3}x^{3/2} require x \geq 0, so the result is stated on [0, \infty). The function is continuous at x = 0, so 0 is a valid limit of integration even though the derivative of \sqrt{x} is undefined there.

The integral of sqrt x is (2/3)x^(3/2) + C. Rewrite sqrt x as x^(1/2) and use the power rule for antiderivatives: add 1 to the exponent to get 3/2, then divide by 3/2, which is the same as multiplying by 2/3. The answer holds for x >= 0, where sqrt x is defined.

\int \sqrt{x}\,dx = \frac{2}{3}x^{3/2} + C

How to integrate sqrt x with the power rule

The radical is the only thing standing in the way. Rewrite $\sqrt{x}$ as a power of $x$ and the problem becomes a routine application of the power rule for antiderivatives.

\int \sqrt{x}\,dx = \int x^{1/2}\,dx

The power rule for antiderivatives raises the exponent by 1 and divides by the new exponent. It works for every exponent except $n = -1$ , where the antiderivative is $\ln|x|$ instead.

\int x^n\,dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1

Here $n = \frac{1}{2}$ , so $n + 1 = \frac{3}{2}$ . Dividing by $\frac{3}{2}$ is the same as multiplying by $\frac{2}{3}$ , which is where the fraction in the answer comes from.

\int x^{1/2}\,dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C

Check by differentiating

Every antiderivative can be verified in one line. Differentiating $\frac{2}{3}x^{3/2}$ brings the $\frac{3}{2}$ down front, and $\frac{2}{3} \cdot \frac{3}{2} = 1$ , leaving exactly the integrand.

\frac{d}{dx}\left[\frac{2}{3}x^{3/2}\right] = \frac{2}{3} \cdot \frac{3}{2}x^{1/2} = \sqrt{x}

The definite integral: area under sqrt x

Once you have the antiderivative, the Fundamental Theorem of Calculus turns any definite integral of $\sqrt{x}$ into arithmetic. Take the area under the curve from $x = 0$ to $x = 4$ .

\int_0^4 \sqrt{x}\,dx = \left[\frac{2}{3}x^{3/2}\right]_0^4

Evaluate $4^{3/2}$ by taking the root first and then the power: $\sqrt{4} = 2$ , and $2^3 = 8$ . That order keeps the numbers small enough to do without a calculator.

\frac{2}{3}(4)^{3/2} - \frac{2}{3}(0)^{3/2} = \frac{2}{3}(8) = \frac{16}{3}

The same antiderivative handles any interval. From $x = 1$ to $x = 9$ , the endpoints give $9^{3/2} = 27$ and $1^{3/2} = 1$ .

\int_1^9 \sqrt{x}\,dx = \frac{2}{3}(27) - \frac{2}{3}(1) = \frac{52}{3}

Zero is a legal endpoint here

$\sqrt{x}$ is continuous on $[0, 4]$ , so $\int_0^4 \sqrt{x}\,dx$ is an ordinary definite integral, not an improper one. The derivative of $\sqrt{x}$ blows up at $x = 0$ , but the function itself does not, and integration only needs the function.

Where this integral shows up on the AP exam

Rewriting radicals as fractional powers is Topic 6.8 (Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation). Unit 6 carries a weighting of 15 to 20 percent on both AB and BC, making it the heaviest single unit on either exam.

The integral rarely appears bare. More often $\sqrt{x}$ is one term inside a longer polynomial-style integrand, or it defines a region whose area or volume you are asked for in Unit 8 (Applications of Integration).

When the radical hides a composite, the power rule alone is not enough and you need u-substitution. With $u = 3x + 1$ you get $du = 3\,dx$ , so the extra factor of $\frac{1}{3}$ combines with the $\frac{2}{3}$ from the power rule.

\int \sqrt{3x+1}\,dx = \frac{2}{9}(3x+1)^{3/2} + C

Average value questions also lean on this antiderivative. The average value of $\sqrt{x}$ on $[0, 4]$ divides the accumulated area by the width of the interval.

\frac{1}{4}\int_0^4 \sqrt{x}\,dx = \frac{4}{3}

Common mistakes

Dividing by the old exponent instead of the new one. That gives $\frac{x^{3/2}}{1/2} = 2x^{3/2}$ , whose derivative is $3\sqrt{x}$ , three times too big.
Running the derivative power rule backwards. Differentiating multiplies by the exponent; antidifferentiating divides by the new exponent. The $\frac{2}{3}$ has to end up in the numerator.
Dropping the constant of integration. An indefinite integral is a family of functions, so $+ C$ is part of the answer and costs a point without it.
Computing $4^{3/2}$ as $4 \cdot \frac{3}{2} = 6$ . The exponent is not a multiplier: take the square root, then cube.
Applying the bare result to $\sqrt{u}$ without accounting for $u'$ . Only $\int \sqrt{x}\,dx$ gives $\frac{2}{3}x^{3/2}$ ; a composite needs substitution first.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

What is the integral of sqrt x?

$\int \sqrt{x}\,dx = \frac{2}{3}x^{3/2} + C$ . Rewrite $\sqrt{x}$ as $x^{1/2}$ , add 1 to the exponent to get $\frac{3}{2}$ , then divide by $\frac{3}{2}$ , which multiplies by $\frac{2}{3}$ .

Why is the answer 2/3 and not 3/2?

The power rule divides by the new exponent, and dividing by $\frac{3}{2}$ is the same as multiplying by $\frac{2}{3}$ . You can confirm the direction by differentiating: $\frac{d}{dx}\left[\frac{2}{3}x^{3/2}\right] = \frac{2}{3} \cdot \frac{3}{2}x^{1/2} = \sqrt{x}$ . If the constant were $\frac{3}{2}$ , the derivative would come out as $\frac{9}{4}\sqrt{x}$ .

What is the integral of sqrt x from 0 to 4?

It is $\frac{16}{3}$ . Evaluate $\frac{2}{3}x^{3/2}$ at the endpoints: $4^{3/2} = 8$ because $\sqrt{4} = 2$ and $2^3 = 8$ , so you get $\frac{2}{3}(8) - 0 = \frac{16}{3}$ .

How do you integrate sqrt(ax + b)?

Use u-substitution, not the bare power rule. With $u = ax + b$ you have $du = a\,dx$ , so $\int \sqrt{ax+b}\,dx = \frac{2}{3a}(ax+b)^{3/2} + C$ . For example $\int \sqrt{3x+1}\,dx = \frac{2}{9}(3x+1)^{3/2} + C$ .

What is the domain of the antiderivative?

Both $\sqrt{x}$ and $\frac{2}{3}x^{3/2}$ require $x \geq 0$ , so the result is stated on $[0, \infty)$ . The function is continuous at $x = 0$ , so $0$ is a valid limit of integration even though the derivative of $\sqrt{x}$ is undefined there.