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r(x)=int_(-2)^(x)2te^(4t^(2))dh

r(x)=∫−2x2te4t2 dhr(x)=\int_{-2}^{x}2te^{4t^{2}}\,dh

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Evaluate definite integrals using the chain ruleright chevron icon

Full solution

Q. r(x)=∫−2x2te4t2 dhr(x)=\int_{-2}^{x}2te^{4t^{2}}\,dh
  1. Identify integral: Identify the integral to be solved.\newlineWe need to solve the integral of 2t⋅e4t22t \cdot e^{4t^2} from −2-2 to xx.
  2. Perform substitution: Perform a substitution to simplify the integral.\newlineLet u=4t2u = 4t^2, then du=8tdtdu = 8t dt, so dt=du8tdt = \frac{du}{8t}.\newlineSubstitute into the integral:\newline∫2t⋅e4t2dt=∫2t⋅eu⋅(du8t)=14∫eudu\int 2t \cdot e^{4t^2} dt = \int 2t \cdot e^u \cdot \left(\frac{du}{8t}\right) = \frac{1}{4} \int e^u du.
  3. Solve with new variable: Solve the integral with the new variable.\newlineThe integral of eue^u with respect to uu is eu+Ce^u + C.\newlineSo, 14∫eu du=14eu+C\frac{1}{4} \int e^u \, du = \frac{1}{4} e^u + C.
  4. Substitute back to t: Substitute back to the original variable t. Since u=4t2u = 4t^2, we have 14⋅e4t2+C\frac{1}{4} \cdot e^{4t^2} + C.
  5. Evaluate definite integral: Evaluate the definite integral from −2-2 to xx.r(x)=(14⋅e4x2−14⋅e4(−2)2).r(x) = \left(\frac{1}{4} \cdot e^{4x^2} - \frac{1}{4} \cdot e^{4(-2)^2}\right).Simplify the expression:r(x)=(14⋅e4x2−14⋅e16).r(x) = \left(\frac{1}{4} \cdot e^{4x^2} - \frac{1}{4} \cdot e^{16}\right).

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