Q.Evaluate $\int_{-1}^1 5x^4 \sqrt{x^5 + 1}\, dx$

The key idea is the Definite Substitution Method — we change the variable and adjust the limits accordingly.

Let $u = x^5 + 1$. Then $du = 5x^4\,dx$, which matches the integrand exactly.

When $x = -1$, $u = (-1)^5 + 1 = 0$.

When $x = 1$, $u = 1^5 + 1 = 2$.

The integral becomes:

Evaluating:

Why substitution works here

When you see an integral of the form $\int f(g(x)) \cdot g'(x)\, dx$, the chain rule in reverse tells you to substitute $u = g(x)$. Here, the integrand is $5x^4 \sqrt{x^5+1}$. Notice that the derivative of $x^5+1$ is $5x^4$ — that’s exactly the factor sitting outside the square root. This is not a coincidence; it’s the hallmark of a function and its derivative appearing together.

The definite integral version of substitution is even cleaner: you change the limits along with the variable, so you never have to “back-substitute.” You just evaluate the new integral in $u$ at the new limits.

Step-by-step solution

1. Choose the substitution.

Let $u = x^5 + 1$. Then the differential is $du = 5x^4\, dx$. That’s precisely the $5x^4\, dx$ part of the integrand.

2. Change the limits of integration.

When $x = -1$:

$u = (-1)^5 + 1 = -1 + 1 = 0$

When $x = 1$:

$u = (1)^5 + 1 = 1 + 1 = 2$

So the integral in $x$ from $-1$ to $1$ becomes an integral in $u$ from $0$ to $2$.

3. Rewrite the integral.

The original integral is:

After substitution, it becomes:

4. Evaluate the $u$-integral.

Recall $\sqrt{u} = u^{1/2}$. Its antiderivative is $\frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}$.

So:

5. Plug in the limits.

At $u=2$: $\frac{2}{3} (2)^{3/2} = \frac{2}{3} \cdot 2\sqrt{2} = \frac{4\sqrt{2}}{3}$

At $u=0$: $\frac{2}{3} (0)^{3/2} = 0$

Subtract: $\frac{4\sqrt{2}}{3} - 0 = \frac{4\sqrt{2}}{3}$

Method: Definite Integral by Substitution (Changing the Limits)

Use this for a definite integral $\int_a^b f(g(x))g'(x)\,dx$ where the derivative of the inner function appears: substitute and change the limits so no back-substitution is required.

Steps

Step 1: Choose $u=g(x)$ where $g'(x)$ is present.

Spot the inner function whose derivative multiplies the rest. For $\int_{-1}^1 5x^4\sqrt{x^5+1}\,dx$, take $u=x^5+1$, since $du=5x^4\,dx$ matches the outside factor.

Step 2: Convert the limits to $u$-values.

Compute $u$ at each endpoint: $x=-1\Rightarrow u=0$, $x=1\Rightarrow u=2$. The integral becomes $\int_0^2\sqrt{u}\,du$.

Step 3: Integrate and evaluate directly in $u$.

Common Mistakes

Mistake 1: Not changing the limits after substituting.

Why it's wrong: keeping $-1$ and $1$ as bounds for a $u$-integral gives the wrong value. Correct approach: recompute the limits as $u(-1)=0$ and $u(1)=2$.

Mistake 2: Back-substituting to $x$ and then using $u$-limits (or vice versa).

Why it's wrong: mixing the two conventions double-counts the change. Correct approach: either convert limits and stay in $u$, or keep $x$-limits and back-substitute — never both.

Mistake 3: Overlooking that $du=5x^4\,dx$ matches exactly.

Why it's wrong: inserting a stray constant when the factor already matches distorts the answer. Correct approach: confirm the outside factor equals $du$ before substituting.