-0.636827
3
5
0.989992
-0.283662
1
1
-0.636827
3
5
0.989992
-0.283662
1
1
The Integration by Substitution Calculator evaluates definite integrals of composite functions of the form $$\int_a^b f(\alpha x + \beta)\,dx$$ using the u-substitution method. This is the integration counterpart of the chain rule for differentiation and is perhaps the most frequently used technique for evaluating integrals in both pure and applied mathematics.
The idea behind u-substitution is elegantly simple. If the integrand contains a composite function $$f(g(x))$$ and the derivative $$g'(x)$$ appears as a factor (or can be accounted for), then setting $$u = g(x)$$ transforms the integral into $$\int f(u)\,du$$ divided by the appropriate constant. For the linear substitution $$u = ax + b$$, we have $$du = a\,dx$$, so $$dx = \frac{1}{a}du$$, and the integral becomes $$\frac{1}{a}\int f(u)\,du$$.
This calculator supports six common outer functions: $$\sin(u)$$, $$\cos(u)$$, $$e^u$$, $$\sqrt{u}$$, $$u^2$$, and $$u^3$$. Each has a known antiderivative, so the substituted integral can be evaluated in closed form. For example, $$\int \sin(2x+3)\,dx = -\frac{1}{2}\cos(2x+3) + C$$ because $$u = 2x+3$$, $$du = 2\,dx$$, and $$\int \sin(u)\,du = -\cos(u)$$.
The substitution method was implicitly used by Newton and Leibniz in the late 17th century, though Euler and the Bernoulli family formalized and extended the technique in the 18th century. Today it is one of the first integration techniques taught in calculus courses worldwide, serving as the gateway to more advanced methods like trigonometric substitution, partial fractions, and integration by parts.
In applications, u-substitution appears constantly. In physics, computing work done by a variable force along a curved path often requires substituting the parameterization. In statistics, transformations of random variables use substitution to derive new probability density functions. In engineering, frequency-domain analysis via Fourier and Laplace transforms relies heavily on substitution. In chemistry, reaction kinetics with temperature-dependent rate constants involve substitutions to evaluate Arrhenius-type integrals.
The calculator also transforms the bounds automatically. When you set $$u = ax + b$$, the original bounds $$x = x_1$$ and $$x = x_2$$ become $$u_1 = ax_1 + b$$ and $$u_2 = ax_2 + b$$. This means you never need to back-substitute: the definite integral is evaluated entirely in terms of $$u$$. Enter your parameters and bounds to see the complete substitution process and result.
Given the integral $$\int_{x_1}^{x_2} f(ax + b)\,dx$$, the substitution $$u = ax + b$$ yields $$du = a\,dx$$, so:
$$\int_{x_1}^{x_2} f(ax+b)\,dx = \frac{1}{a}\int_{u_1}^{u_2} f(u)\,du = \frac{1}{a}\left[F(u)\right]_{u_1}^{u_2}$$
where $$u_1 = ax_1 + b$$, $$u_2 = ax_2 + b$$, and $$F(u)$$ is the antiderivative of $$f(u)$$.
Antiderivatives used:
$$f(u) = \sin(u) \Rightarrow F(u) = -\cos(u)$$
$$f(u) = \cos(u) \Rightarrow F(u) = \sin(u)$$
$$f(u) = e^u \Rightarrow F(u) = e^u$$
$$f(u) = \sqrt{u} \Rightarrow F(u) = \frac{2}{3}u^{3/2}$$
$$f(u) = u^2 \Rightarrow F(u) = \frac{u^3}{3}$$
$$f(u) = u^3 \Rightarrow F(u) = \frac{u^4}{4}$$
The final result is $$\frac{1}{a}\bigl(F(u_2) - F(u_1)\bigr)$$.
The Definite Integral Result gives the net signed area under the curve $$f(ax+b)$$ between the specified bounds. This accounts for the chain rule factor $$\frac{1}{a}$$ automatically.
The u at Lower/Upper Bound values show the transformed limits of integration. These are computed as $$u = ax + b$$ at each original bound. When $$a > 0$$, the order of limits is preserved; when $$a < 0$$, the limits reverse, which the formula handles correctly via the $$\frac{1}{a}$$ factor (negative $$a$$ flips the sign).
The Antiderivative F(u) values show $$F(u)$$ evaluated at the transformed upper and lower bounds. The difference $$F(u_2) - F(u_1)$$, scaled by $$\frac{1}{a}$$, gives the final answer.
Inputs
Results
Substitution u = 2x+3. At x=0: u=3; at x=pi/2: u=3+pi=6.1416. F(u) = -cos(u). F(6.1416) = -cos(6.1416) = 0.8494. F(3) = -cos(3) = 0.9899. Result = (1/2)(0.8494 - 0.9899) = (1/2)(-0.1405) = -0.0702.
Inputs
Results
Substitution u = 3x+1. At x=0: u=1; at x=2: u=7. F(u) = u^3/3. F(7) = 343/3 = 114.333. F(1) = 1/3 = 0.333. Result = (1/3)(114.333 - 0.333) = (1/3)(114) = 38. Alternatively: expand (3x+1)^2 = 9x^2+6x+1, integrate to get 3x^3+3x^2+x, evaluate at 2: 24+12+2 = 38.
U-substitution (also called the substitution rule) is the reverse of the chain rule for differentiation. If you can identify a composite function $$f(g(x))$$ in the integrand and the inner derivative $$g'(x)$$ is present (or is a constant), you substitute $$u = g(x)$$ to simplify the integral into $$\int f(u)\,du$$ up to a constant factor.
When $$u = ax + b$$, differentiating gives $$du = a\,dx$$, so $$dx = \frac{1}{a}du$$. This factor of $$\frac{1}{a}$$ compensates for the chain rule. It means the integral of $$f(ax+b)$$ is $$\frac{1}{a}$$ times the integral of $$f(u)$$.
This calculator handles only linear inner functions $$u = ax + b$$. For nonlinear substitutions like $$u = x^2 + 1$$ or $$u = \sin(x)$$, the factor $$g'(x)$$ is not constant, so the integrand must contain $$g'(x)$$ as an explicit factor. Those cases require a more general substitution calculator.
A negative $$a$$ reverses the direction of the substitution. The transformed limits swap order, but dividing by the negative $$a$$ corrects the sign automatically. The formula $$\frac{1}{a}[F(u)]_{u_1}^{u_2}$$ handles this correctly without any special cases.
The square root function $$\sqrt{u}$$ requires $$u \geq 0$$ in real analysis. If the transformed bounds produce negative $$u$$ values, the result may not be meaningful in the real number system. Ensure that $$ax + b \geq 0$$ over your entire integration interval when using the sqrt option.
Expand the composite function if possible (e.g., $$(ax+b)^2 = a^2x^2 + 2abx + b^2$$), integrate term by term, and evaluate at the bounds. For trigonometric and exponential functions, use a table of integrals or differentiate the antiderivative to confirm it matches the original integrand.
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